Solution

The area of a circle ( $A$ ) with radius $(r)$ is given by,

$A=\pi r^{2}$

Now, the rate of change of the area with respect to its radius is given by,

$\frac{d A}{d r}=\frac{d}{d r}(\pi r^{2})=2 \pi r$

1. When $r=3 cm$,

$\frac{d A}{d r}=2 \pi(3)=6 \pi$

Hence, the area of the circle is changing at the rate of $6 cm^{2} / s$ when its radius is $3 cm$.

2. When $r=4 cm$,

$\frac{d A}{d r}=2 \pi(4)=8 \pi$

Hence, the area of the circle is changing at the rate of $8 cm^{2} / s$ when its radius is $4 cm$.

Solution

Let $x$ be the length of a side, $V$ be the volume, and $s$ be the surface area of the cube.

Then, $V=x^{3}$ and $S=6 x^{2}$ where $x$ is a function of time $t$.

It is given that $\frac{d V}{d t}=8 cm^{3} / s$.

Then, by using the chain rule, we have:

$ \therefore=\frac{d V}{d t}=\frac{d}{d t}(x^{3})=\frac{d}{d x}(x^{3}) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t} $

$ \begin{equation*} \Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}} \tag{1} \end{equation*} $

$ \begin{aligned} & \text{ Now, } \frac{d S}{d t}=\frac{d}{d t}(6 x^{2})=\frac{d}{d x}(6 x^{2}) \cdot \frac{d x}{d t} \quad \text{ [By chain rule] } \\ & =12 x \cdot \frac{d x}{d t}=12 x \cdot(\frac{8}{3 x^{2}})=\frac{32}{x} \end{aligned} $

Hence, if the length of the edge of the cube is $12 cm$, then the surface area is increasing

at the rate of $\frac{8}{3} cm^{2} / s$.

Solution

The area of a circle $(A)$ with radius $(r)$ is given by,

$A=\pi r^{2}$

Now, the rate of change of area $(A)$ with respect to time $(t)$ is given by,

$\frac{d A}{d t}=\frac{d}{d t}(\pi r^{2}) \cdot \frac{d r}{d t}=2 \pi r \frac{d r}{d t} \quad$ [By chain rule]

It is given that,

$\frac{d r}{d t}=3 cm / s$ $\therefore \frac{d A}{d t}=2 \pi r(3)=6 \pi r$

Thus, when $r=10 cm$,

$\frac{d A}{d t}=6 \pi(10)=60 \pi cm^{2} / s$

Hence, the rate at which the area of the circle is increasing when the radius is $10 cm$ is $60 п cm^{2} / s$.

Solution

Let $x$ be the length of a side and $V$ be the volume of the cube. Then,

$V=x^{3}$.

$\therefore \frac{d V}{d t}=3 x^{2} \cdot \frac{d x}{d t}$ (By chain rule)

It is given that,

$\frac{d x}{d t}=3 cm / s$

$\therefore \frac{d V}{d t}=3 x^{2}(3)=9 x^{2}$

Thus, when $x=10 cm$,

$\frac{d V}{d t}=9(10)^{2}=900 cm^{3} / s$

Hence, the volume of the cube is increasing at the rate of $900 cm^{3} / s$ when the edge is 10 $cm$ long.

Solution

The area of a circle (A) with radius $(r)$ is given by $A=\pi r^{2}$.

Therefore, the rate of change of area $(A)$ with respect to time $(t)$ is given by,

$\frac{d A}{d t}=\frac{d}{d t}(\pi r^{2})=\frac{d}{d r}(\pi r^{2}) \frac{d r}{d t}=2 \pi r \frac{d r}{d t}$ [By chain rule]

It is given that $\frac{d r}{d t}=5 cm / s$.

Thus, when $r=8 cm$,

$\frac{d A}{d t}=2 \pi(8)(5)=80 \pi$

Hence, when the radius of the circular wave is $8 cm$, the enclosed area is increasing at the rate of $80 \pi cm^{2} / s$.

Solution

The circumference of a circle $(C)$ with radius $(r)$ is given by $C=2 \pi r$.

Therefore, the rate of change of circumference (C) with respect to time $(t)$ is given by, $\frac{d C}{d t}=\frac{d C}{d r} \cdot \frac{d r}{d t}$ (By chain rule) $=\frac{d}{d r}(2 \pi r) \frac{d r}{d t}$

$=2 \pi \cdot \frac{d r}{d t}$

It is given that $\frac{d r}{d t}=0.7 cm / s$.

Hence, the rate of increase of the circumference is $2 \pi(0.7)=1.4 \pi cm / s$.

Solution

Since the length $(x)$ is decreasing at the rate of $5 cm /$ minute and the width $(y)$ is increasing at the rate of $4 cm /$ minute, we have:

$\frac{d x}{d t}=-5 cm / min$ and $\frac{d y}{d t}=4 cm / min$

(a) The perimeter $(P)$ of a rectangle is given by,

$P=2(x+y)$

$\therefore \frac{d P}{d t}=2(\frac{d x}{d t}+\frac{d y}{d t})=2(-5+4)=-2 cm / min$

Hence, the perimeter is decreasing at the rate of $2 cm / min$.

(b) The area (A) of a rectangle is given by,

$A=x \times y$

$\therefore \frac{d A}{d t}=\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t}=-5 y+4 x$

When $x=8 cm$ and $y=6 cm, \frac{d A}{d t}=(-5 \times 6+4 \times 8) cm^{2} / min=2 cm^{2} / min$

Hence, the area of the rectangle is increasing at the rate of $2 cm^{2} / min$.

Solution

The volume of a sphere $(V)$ with radius $(r)$ is given by,

$V=\frac{4}{3} \pi r^{3}$

$\therefore$ Rate of change of volume $(V)$ with respect to time $(t)$ is given by,

$\frac{d V}{d t}=\frac{d V}{d r} \cdot \frac{d r}{d t} _{\text{[By chain rule] }}$

$=\frac{d}{d r}(\frac{4}{3} \pi r^{3}) \cdot \frac{d r}{d t}$

$=4 \pi r^{2} \cdot \frac{d r}{d t}$

It is given that $\frac{d V}{d t}=900 cm^{3} / s$.

$\therefore 900=4 \pi r^{2} \cdot \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{900}{4 \pi r^{2}}=\frac{225}{\pi r^{2}}$

Therefore, when radius $=15 cm$,

$\frac{d r}{d t}=\frac{225}{\pi(15)^{2}}=\frac{1}{\pi}$

Hence, the rate at which the radius of the balloon increases when the radius is $15 cm$

$1 cm / s$.

is $\pi$

Solution

The volume of a sphere $(V)$ with radius $(r)$ is given by $V=\frac{4}{3} \pi r^{3}$.

Rate of change of volume $(V)$ with respect to its radius $(r)$ is given by,

$\frac{d V}{d r}=\frac{d}{d r}(\frac{4}{3} \pi r^{3})=\frac{4}{3} \pi(3 r^{2})=4 \pi r^{2}$

Therefore, when radius $=10 cm$,

$\frac{d V}{d r}=4 \pi(10)^{2}=400 \pi$

Hence, the volume of the balloon is increasing at the rate of $400 cm^{3} / s$.

Solution

Let $y m$ be the height of the wall at which the ladder touches. Also, let the foot of the ladder be $x$ maway from the wall.

Then, by Pythagoras theorem, we have:

$x^{2}+y^{2}=25$ [Length of the ladder $=5 m$ ]

$\Rightarrow y=\sqrt{25-x^{2}}$

Then, the rate of change of height $(y)$ with respect to time $(t)$ is given by,

$\frac{d y}{d t}=\frac{-x}{\sqrt{25-x^{2}}} \cdot \frac{d x}{d t}$

It is given that $\frac{d x}{d t}=2 cm / s$. $\therefore \frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^{2}}}$

Now, when $x=4 m$, we have:

$\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^{2}}}=-\frac{8}{3}$

Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{8}{3} cm / s$.

Solution

The equation of the curve is given as:

$6 y=x^{3}+2$

The rate of change of the position of the particle with respect to time $(t)$ is given by,

$6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0$

$\Rightarrow 2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}$

When the $y$-coordinate of the particle changes 8 times as fast as the

$x$-coordinate i.e., $(\frac{d y}{d t}=8 \frac{d x}{d t})$, we have:

$2(8 \frac{d x}{d t})=x^{2} \frac{d x}{d t}$

$\Rightarrow 16 \frac{d x}{d t}=x^{2} \frac{d x}{d t}$

$\Rightarrow(x^{2}-16) \frac{d x}{d t}=0$

$\Rightarrow x^{2}=16$

$\Rightarrow x= \pm 4$

When $x=4, y=\frac{4^{3}+2}{6}=\frac{66}{6}=11$.

When $x=-4, y=\frac{(-4)^{3}+2}{6}=-\frac{62}{6}=-\frac{31}{3}$.

Hence, the points required on the curve are $(4,11)$ and $(-4, \frac{-31}{3})$.

Solution

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble $(V)$ with radius $(r)$ is given by,

$V=\frac{4}{3} \pi r^{3}$

The rate of change of volume $(V)$ with respect to time $(t)$ is given by,

$ \begin{matrix} \frac{d V}{d t} & =\frac{4}{3} \pi \frac{d}{d r}(r^{3}) \cdot \frac{d r}{d t} & \text{ [By chain rule] } \\ & =\frac{4}{3} \pi(3 r^{2}) \frac{d r}{d t} & \\ & =4 \pi r^{2} \frac{d r}{d t} & \end{matrix} $

It is given that $\frac{d r}{d t}=\frac{1}{2} cm / s$.

Therefore, when $r=1 cm$,

$\frac{d V}{d t}=4 \pi(1)^{2}(\frac{1}{2})=2 \pi cm^{3} / s$

Hence, the rate at which the volume of the bubble increases is $2 \pi cm^{3} / s$.

Solution

The volume of a sphere $(V)$ with radius $(r)$ is given by,

$V=\frac{4}{3} \pi r^{3}$

It is given that:

Diameter $=\frac{3}{2}(2 x+1)$

$\Rightarrow r=\frac{3}{4}(2 x+1)$

$\therefore V=\frac{4}{3} \pi(\frac{3}{4})^{3}(2 x+1)^{3}=\frac{9}{16} \pi(2 x+1)^{3}$

Hence, the rate of change of volume with respect to $x$ is as

$ \frac{d V}{d x}=\frac{9}{16} \pi \frac{d}{d x}(2 x+1)^{3}=\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2=\frac{27}{8} \pi(2 x+1)^{2} . $

Solution

The volume of a cone $(V)$ with radius $(r)$ and height $(h)$ is given by,

$V=\frac{1}{3} \pi r^{2} h$

It is given that,

$h=\frac{1}{6} r \Rightarrow r=6 h$

$\therefore V=\frac{1}{3} \pi(6 h)^{2} h=12 \pi h^{3}$

The rate of change of volume with respect to time $(t)$ is given by,

$\frac{d V}{d t}=12 \pi \frac{d}{d h}(h^{3}) \cdot \frac{d h}{d t}$ [By chain rule]

$=12 \pi(3 h^{2}) \frac{d h}{d t}$

$=36 \pi h^{2} \frac{d h}{d t}$

It is also given that $\frac{d V}{d t}=12 cm^{3} / s$.

Therefore, when $h=4 cm$, we have:

$ \begin{aligned} & 12=36 \pi(4)^{2} \frac{d h}{d t} \\ & \Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi(16)}=\frac{1}{48 \pi} \end{aligned} $

Hence, when the height of the sand cone is $4 cm$, its height is increasing at the rate

of $\frac{1}{48 \pi} cm / s$.

Solution

Marginal cost is the rate of change of total cost with respect to output.

$\therefore$ Marginal cost (MC) $=\frac{d C}{d x}=0.007(3 x^{2})-0.003(2 x)+15$

$=0.021 x^{2}-0.006 x+15$

When $x=17, M C=0.021(17^{2})-0.006(17)+15$ $=0.021(289)-0.006(17)+15$

$=6.069-0.102+15$

$=20.967$

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

Solution

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

$\therefore$ Marginal Revenue (MR) $=\frac{d R}{d x}=13(2 x)+26=26 x+26$

When $x=7$,

$M R=26(7)+26=182+26=208$

Hence, the required marginal revenue is Rs 208.

Solution

The area of a circle ( $A$ ) with radius $(r)$ is given by,

$A=\pi r^{2}$

Therefore, the rate of change of the area with respect to its radius $r$ is

$\frac{d A}{d r}=\frac{d}{d r}(\pi r^{2})=2 \pi r$. $\therefore$ When $r=6 cm$,

$\frac{d A}{d r}=2 \pi \times 6=12 \pi cm^{2} / s$

Hence, the required rate of change of the area of a circle is $12 \pi cm^{2} / s$.

The correct answer is $B$.

Solution

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

$\therefore$ Marginal Revenue (MR) $=\frac{d R}{d x}=3(2 x)+36=6 x+36$

$\therefore$ When $x=15$,

$M R=6(15)+36=90+36=126$

Hence, the required marginal revenue is Rs 126 .

The correct answer is D.

Solution

Let ${ }^{x_1}$ and $x_2$ be any two numbers in $\mathbf{R}$.

Then, we have:

$x_1<x_2 \Rightarrow 3 x_1<3 x_2 \Rightarrow 3 x_1+17<3 x_2+17 \Rightarrow f(x_1)<f(x_2)$

Hence, $f$ is strictly increasing on $\mathbf{R}$.

Alternate method:

$f(x)=3>0$, in every interval of $\mathbf{R}$.

Thus, the function is strictly increasing on $\mathbf{R}$.

Solution

Let ${ }^{x_1}$ and $x_2$ be any two numbers in $\mathbf{R}$.

Then, we have:

$x_1<x_2 \Rightarrow 2 x_1<2 x_2 \Rightarrow e^{2 x_1}<e^{2 x_2} \Rightarrow f(x_1)<f(x_2)$

Hence, $f$ is strictly increasing on $\mathbf{R}$.

Solution

The given function is $f(x)=\sin x$.

$\therefore f^{\prime}(x)=\cos x$ (a) Since for each $x \in(0, \frac{\pi}{2}), \cos x>0$, we have $f^{\prime}(x)>0$.

Hence, $f$ is strictly increasing in $(0, \frac{\pi}{2})$.

(b) Since for each $x \in(\frac{\pi}{2}, \pi), \cos x<0$, we have $f^{\prime}(x)<0$.

Hence, $f$ is strictly decreasing in $(\frac{\pi}{2}, \pi)$.

(c) From the results obtained in (a) and (b), it is clear that $f$ is neither increasing nor decreasing in $(0, \pi)$.

Solution

The given function is $f(x)=2 x^{2}-3 x$.

$f^{\prime}(x)=4 x-3$

$\therefore f^{\prime}(x)=0 \Rightarrow x=\frac{3}{4}$

Now, the point $\frac{3}{4}$ divides the real line into two disjoint intervals i.e., $(-\infty, \frac{3}{4})$ and $(\frac{3}{4}, \infty)$.

In interval $(-\infty, \frac{3}{4}), f^{\prime}(x)=4 x-3<0$.

Hence, the given function $(f)$ is strictly decreasing in interval $(-\infty, \frac{3}{4})$.

In interval $(\frac{3}{4}, \infty), f^{\prime}(x)=4 x-3>0$.

Hence, the given function $(f)$ is strictly increasing in interval $(\frac{3}{4}, \infty)$.

Solution

The given function is $f(x)=2 x^{3}-3 x^{2}-36 x+7$.

$f^{\prime}(x)=6 x^{2}-6 x-36=6(x^{2}-x-6)=6(x+2)(x-3)$

$\therefore f^{\prime}(x)=0 \Rightarrow x=-2,3$

The points $x=-2$ and $x=3$ divide the real line into three disjoint intervals i.e., $(-\infty,-2),(-2,3)$, and $(3, \infty)$.

In intervals $(-\infty,-2)$ and $(3, \infty), f^{\prime}(x)$ is positive while in interval

$(-2,3), f^{\prime}(x)$ is negative.

Hence, the given function $(f)$ is strictly increasing in intervals

$(-\infty,-2)$ and $(3, \infty)$, while function $(f)$ is strictly decreasing in interval

$(-2,3)$.

Solution

(a) We have,

$f(x)=x^{2}+2 x-5$

$\therefore f^{\prime}(x)=2 x+2$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1$

Point $x=-1$ divides the real line into two disjoint intervals i.e., $(-\infty,-1)$ and $(-1, \infty)$.

In interval $(-\infty,-1), f^{\prime}(x)=2 x+2<0$.

$\therefore f$ is strictly decreasing in interval $(-\infty,-1)$.

Thus, $f$ is strictly decreasing for $x<-1$.

In interval $(-1, \infty), f^{\prime}(x)=2 x+2>0$.

$\therefore f$ is strictly increasing in interval $(-1, \infty)$.

Thus, $f$ is strictly increasing for $x>-1$.

(b) We have,

$f(x)=10-6 x-2 x^{2}$

$\therefore f^{\prime}(x)=-6-4 x$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-\frac{3}{2}$

The point $x=-\frac{3}{2}$ divides the real line into two disjoint intervals

i.e., $(-\infty,-\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$.

In interval $(-\infty,-\frac{3}{2})$ i.e., when $x<-\frac{3}{2}, f^{\prime}(x)=-6-4 x<0$.

$\therefore f$ is strictly increasing for $x<-\frac{3}{2}$.

In interval $(-\frac{3}{2}, \infty)$ i.e., when $x>-\frac{3}{2}, f^{\prime}(x)=-6-4 x<0$.

$\therefore f$ is strictly decreasing for $x>-\frac{3}{2}$.

(c) We have,

$f(x)=-2 x^{3}-9 x^{2}-12 x+1$

$\therefore f^{\prime}(x)=-6 x^{2}-18 x-12=-6(x^{2}+3 x+2)=-6(x+1)(x+2)$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1$ and $x=-2$

Points $x=-1$ and $x=-2$ divide the real line into three disjoint intervals

i.e., $(-\infty,-2),(-2,-1)$, and $(-1, \infty)$.

In intervals $(-\infty,-2)$ and $(-1, \infty)$ i.e., when $x<-2$ and $x>-1$,

$f^{\prime}(x)=-6(x+1)(x+2)<0$. $\therefore f$ is strictly decreasing for $x<-2$ and $x>-1$.

Now, in interval $(-2,-1)$ i.e., when $-2<x<-1, f^{\prime}(x)=-6(x+1)(x+2)>0$.

$\therefore f$ is strictly increasing for $-2<x<-1$.

(d) We have,

$ \begin{aligned} & f(x)=6-9 x-x^{2} \\ & \therefore f^{\prime}(x)=-9-2 x \end{aligned} $

Now, $f^{\prime}$

$(x)=0$ gives $x=-\frac{9}{2}$

The point $x=-\frac{9}{2}$ divides the real line into two disjoint intervals i.e.,

$(-\infty,-\frac{9}{2})$ and $(-\frac{9}{2}, \infty)$.

In interval $(-\infty,-\frac{9}{2})$ i.e., for $x<-\frac{9}{2}, f^{\prime}(x)=-9-2 x>0$.

$\therefore f$ is strictly increasing for $x<-\frac{9}{2}$.

In interval $(-\frac{9}{2}, \infty)$ i.e., for $x>-\frac{9}{2}, f^{\prime}(x)=-9-2 x<0$. $\therefore f$ is strictly decreasing for $x>-\frac{9}{2}$.

(e) We have,

$ \begin{aligned} f(x)= & (x+1)^{3}(x-3)^{3} \\ f^{\prime}(x) & =3(x+1)^{2}(x-3)^{3}+3(x-3)^{2}(x+1)^{3} \\ & =3(x+1)^{2}(x-3)^{2}[x-3+x+1] \\ & =3(x+1)^{2}(x-3)^{2}(2 x-2) \\ & =6(x+1)^{2}(x-3)^{2}(x-1) \end{aligned} $

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1,3,1$

The points $x=-1, x=1$, and $x=3$ divide the real line into four disjoint intervals

i.e., $(-\infty,-1),(-1,1),(1,3)$, and $(3, \infty)$.

In intervals $(-\infty,-1)$ and $(-1,1), f^{\prime}(x)=6(x+1)^{2}(x-3)^{2}(x-1)<0$.

$\therefore f$ is strictly decreasing in intervals $(-\infty,-1)$ and $(-1,1)$.

In intervals $(1,3)$ and $(3, \infty), f^{\prime}(x)=6(x+1)^{2}(x-3)^{2}(x-1)>0$.

$\therefore f$ is strictly increasing in intervals $(1,3)$ and $(3, \infty)$.

Solution

We have,

$ \begin{aligned} & y=\log (1+x)-\frac{2 x}{2+x} \\ & \therefore \frac{d y}{d x}=\frac{1}{1+x}-\frac{(2+x)(2)-2 x(1)}{(2+x)^{2}}=\frac{1}{1+x}-\frac{4}{(2+x)^{2}}=\frac{x^{2}}{(2+x)^{2}} \end{aligned} $

Now, $\frac{d y}{d x}=0$

$\Rightarrow \frac{x^{2}}{(2+x)^{2}}=0$

$\Rightarrow x^{2}=0 \quad[(2+x) \neq 0$ as $x>-1]$

$\Rightarrow x=0$

Since $x>-1$, point $x=0$ divides the domain $(-1, \infty)$ in two disjoint intervals i.e., $-1<$ $x<0$ and $x>0$.

When $-1<x<0$, we have:

$x<0 \Rightarrow x^{2}>0$

$x>-1 \Rightarrow(2+x)>0 \Rightarrow(2+x)^{2}>0$

$\therefore y^{\prime}=\frac{x^{2}}{(2+x)^{2}}>0$

Also, when $x>0$ :

$x>0 \Rightarrow x^{2}>0,(2+x)^{2}>0$

$\therefore y^{\prime}=\frac{x^{2}}{(2+x)^{2}}>0$

Hence, function $f$ is increasing throughout this domain.

Solution

We have,

$y=[x(x-2)]^{2}=[x^{2}-2 x]^{2}$

$\therefore \frac{d y}{d x}=y^{\prime}=2(x^{2}-2 x)(2 x-2)=4 x(x-2)(x-1)$

$\therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1$.

The points $x=0, x=1$, and $x=2$ divide the real line into four disjoint intervals i.e., $(-\infty, 0),(0,1)(1,2)$, and $(2, \infty)$.

In intervals $(-\infty, 0)$ and $(1,2), \frac{d y}{d x}<0$.

$\therefore y$ is strictly decreasing in intervals $(-\infty, 0)$ and $(1,2)$.

However, in intervals $(0,1)$ and $(2, \infty), \frac{d y}{d x}>0$.

$\therefore y$ is strictly increasing in intervals $(0,1)$ and $(2, \infty)$.

$\therefore y$ is strictly increasing for $0<x<1$ and $x>2$.

Solution

We have,

$ \begin{aligned} & y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta \\ & \begin{aligned} \therefore \frac{d y}{d x} & =\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1 \\ \quad & \frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1 \\ \quad & \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1 \end{aligned} \end{aligned} $

Now, $\frac{d y}{d x}=0$.

$\Rightarrow \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}=1$

$\Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta$

$\Rightarrow \cos ^{2} \theta-4 \cos \theta=0$

$\Rightarrow \cos \theta(\cos \theta-4)=0$

$\Rightarrow \cos \theta=0$ or $\cos \theta=4$

Since $\cos \theta \neq 4, \cos \theta=0$.

$\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}$

Now,

$\frac{d y}{d x}=\frac{8 \cos \theta+4-(4+\cos ^{2} \theta+4 \cos \theta)}{(2+\cos \theta)^{2}}=\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}$

In interval $(0, \frac{\pi}{2})$, we have $\cos \theta>0$. Also, $4>\cos \theta \Rightarrow 4-\cos \theta>0$.

$\therefore \cos \theta(4-\cos \theta)>0$ and also $(2+\cos \theta)^{2}>0$

$\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0$

$\Rightarrow \frac{d y}{d x}>0$

Therefore, $y$ is strictly increasing in interval $(0, \frac{\pi}{2})$.

Also, the given function is continuous at $x=0$ and $x=\frac{\pi}{2}$.

Hence, $y$ is increasing in interval $[0, \frac{\pi}{2}]$.

Solution

The given function is $f(x)=\log x$.

$\therefore f^{\prime}(x)=\frac{1}{x}$

It is clear that for $x>0, f^{\prime}(x)=\frac{1}{x}>0$.

Hence, $f(x)=\log x$ is strictly increasing in interval $(0, \infty)$.

Solution

The given function is $f(x)=x^{2}-x+1$.

$\therefore f^{\prime}(x)=2 x-1$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$.

The point $\frac{1}{2}$ divides the interval $(-1,1)$ into two disjoint intervals

i.e., $(-1, \frac{1}{2})$ and $(\frac{1}{2}, 1)$.

Now, in interval $(-1, \frac{1}{2}), f^{\prime}(x)=2 x-1<0$.

Therefore, $f$ is strictly decreasing in interval $(-1, \frac{1}{2})$.

However, in interval $(\frac{1}{2}, 1), f^{\prime}(x)=2 x-1>0$.

Therefore, $f$ is strictly increasing in interval $(\frac{1}{2}, 1)$.

Hence, $f$ is neither strictly increasing nor decreasing in interval $(-1,1)$.

Solution

(A) Let $f_1(x)=\cos x$.

$\therefore f_1^{\prime}(x)=-\sin x$

In interval $(0, \frac{\pi}{2}), f_1^{\prime}(x)=-\sin x<0$.

$\therefore f_1(x)=\cos x _{\text{is strictly decreasing in interval }}(0, \frac{\pi}{2})$.

(B) Let $f_2(x)=\cos 2 x$.

$\therefore f_2^{\prime}(x)=-2 \sin 2 x$

Now, $0<x<\frac{\pi}{2} \Rightarrow 0<2 x<\pi \Rightarrow \sin 2 x>0 \Rightarrow-2 \sin 2 x<0$

$\therefore f_2^{\prime}(x)=-2 \sin 2 x<0$ on $(0, \frac{\pi}{2})$

$\therefore f_2(x)=\cos 2 x$ is strictly decreasing in interval $(0, \frac{\pi}{2})$.

(C) Let $f_3(x)=\cos 3 x$.

$\therefore f_3^{\prime}(x)=-3 \sin 3 x$

Now, $f_3^{\prime}(x)=0$.

$\Rightarrow \sin 3 x=0 \Rightarrow 3 x=\pi$, as $x \in(0, \frac{\pi}{2})$

$\Rightarrow x=\frac{\pi}{3}$

The point $x=\frac{\pi}{3}$ divides the interval $(0, \frac{\pi}{2})$ into two disjoint intervals

i.e., $0(0, \frac{\pi}{3})$ and $(\frac{\pi}{3}, \frac{\pi}{2})$.

Now, in interval $(0, \frac{\pi}{3}), f_3(x)=-3 \sin 3 x<0[.$ as $.0<x<\frac{\pi}{3} \Rightarrow 0<3 x<\pi]$.

$\therefore f_3$ is strictly decreasing in interval $(0, \frac{\pi}{3})$.

However, in interval $(\frac{\pi}{3}, \frac{\pi}{2}), f_3(x)=-3 \sin 3 x>0[.$ as $.\frac{\pi}{3}<x<\frac{\pi}{2} \Rightarrow \pi<3 x<\frac{3 \pi}{2}]$. $\therefore f_3$ is strictly increasing in interval $(\frac{\pi}{3}, \frac{\pi}{2})$.

Hence, $f_3$ is neither increasing nor decreasing in interval $(0, \frac{\pi}{2})$.

(D) Let $f_4(x)=\tan x$.

$\therefore f_4^{\prime}(x)=\sec ^{2} x$

In interval $(0, \frac{\pi}{2}), f_4^{\prime}(x)=\sec ^{2} x>0$.

$\therefore f_4$ is strictly increasing in interval $(0, \frac{\pi}{2})$.

Therefore, functions $\cos x$ and $\cos 2 x$ are strictly decreasing in $(0, \frac{\pi}{2})$. Hence, the correct answers are A and B.

Solution

We have, $f(x)=x^{100}+\sin x-1$

$\therefore f^{\prime}(x)=100 x^{99}+\cos x$

In interval $(0,1), \cos x>0$ and $100 x^{99}>0$.

$\therefore f^{\prime}(x)>0$.

Thus, function $f$ is strictly increasing in interval $(0,1)$.

In interval $(\frac{\pi}{2}, \pi), \cos x<0$ and $100 x^{99}>0$. Also, $100 x^{99}>\cos x$.

$\therefore f^{\prime}(x)>0$ in $(\frac{\pi}{2}, \pi)$.

Thus, function $f$ is strictly increasing in interval $(\frac{\pi}{2}, \pi)$.

In interval $(0, \frac{\pi}{2}), \cos x>0$ and $100 x^{99}>0$.

$\therefore 100 x^{99}+\cos x>0$

$\Rightarrow f^{\prime}(x)>0$ on $(0, \frac{\pi}{2})$

$\therefore f$ is strictly increasing in interval $(0, \frac{\pi}{2})$.

Hence, function $f$ is strictly decreasing in none of the intervals.

The correct answer is D.

Solution

We have, $f(x)=x^{2}+a x+1$

$\therefore f^{\prime}(x)=2 x+a$

Now, function $f$ will be increasing in $(1,2)$, if $f^{\prime}(x)>0$ in $(1,2)$.

$f^{\prime}(x)>0$

$\Rightarrow 2 x+a>0$

$\Rightarrow 2 x>-a$

$\Rightarrow \quad x>\frac{-a}{2}$

Therefore, we have to find the least value of $a$ such that

$x>\frac{-a}{2}$, when $x \in(1,2)$.

$\Rightarrow x>\frac{-a}{2}($ when $1<x<2)$

Thus, the least value of $a$ for $f$ to be increasing on $(1,2)$ is given by,

$\frac{-a}{2}=1$

$\frac{-a}{2}=1 \Rightarrow a=-2$

Hence, the required value of $a$ is -2 .

Solution

We have,

$f(x)=x+\frac{1}{x}$

$\therefore f^{\prime}(x)=1-\frac{1}{x^{2}}$

Now,

$f^{\prime}(x)=0 \Rightarrow \frac{1}{x^{2}}=1 \Rightarrow x= \pm 1$

The points $x=1$ and $x=-1$ divide the real line in three disjoint intervals i.e.,

$(-\infty,-1),(-1,1)$, and $(1, \infty)$.

In interval $(-1,1)$, it is observed that:

$-1<x<1$

$\Rightarrow x^{2}<1$

$\Rightarrow 1<\frac{1}{x^{2}}, x \neq 0$

$\Rightarrow 1-\frac{1}{x^{2}}<0, x \neq 0$

$\therefore f^{\prime}(x)=1-\frac{1}{x^{2}}<0$ on $(-1,1) \sim{0}$.

$\therefore f$ is strictly decreasing on $(-1,1) \sim{0}$.

In intervals $(-\infty,-1)$ and $(1, \infty)$, it is observed that: $x<-1$ or $1<x$

$\Rightarrow x^{2}>1$

$\Rightarrow 1>\frac{1}{x^{2}}$

$\Rightarrow 1-\frac{1}{x^{2}}>0$

$\therefore f^{\prime}(x)=1-\frac{1}{x^{2}}>0$ on $(-\infty,-1)$ and $(1, \infty)$.

$\therefore f$ is strictly increasing on $(-\infty, 1)$ and $(1, \infty)$.

Hence, function $f$ is strictly increasing in interval $\mathbf{I}$ disjoint from $(-1,1)$. Hence, the given result is proved.

Solution

We have,

$f(x)=\log \sin x$

$\therefore f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x$

In interval $(0, \frac{\pi}{2}), f^{\prime}(x)=\cot x>0$.

$\therefore f$ is strictly increasing in $(0, \frac{\pi}{2})$.

In interval $(\frac{\pi}{2}, \pi), f^{\prime}(x)=\cot x<0$.

$\therefore f$ is strictly decreasing in $(\frac{\pi}{2}, \pi)$.

Solution

We have,

$f(x)=\log \cos x$

$\therefore f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$

In interval $(0, \frac{\pi}{2}), \tan x>0 \Rightarrow-\tan x<0$.

$\therefore f^{\prime}(x)<0$ on $(0, \frac{\pi}{2})$

$\therefore f$ is strictly decreasing on $(0, \frac{\pi}{2})$.

In interval $(\frac{\pi}{2}, \pi), \tan x<0 \Rightarrow-\tan x>0$.

$\therefore f^{\prime}(x)>0$ on $(\frac{\pi}{2}, \pi)$ $\therefore f$ is strictly increasing on $(\frac{\pi}{2}, \pi)$.

Solution

We have,

$f(x)=x^{3}-3 x^{2}+3 x-100$

$ \begin{aligned} f^{\prime}(x) & =3 x^{2}-6 x+3 \\ & =3(x^{2}-2 x+1) \\ & =3(x-1)^{2} \end{aligned} $

For any $x \in \mathbf{R},(x-1)^{2}>0$.

Thus, $f^{\prime}(x)$ is always positive in $\mathbf{R}$.

Hence, the given function $(f)$ is increasing in $\mathbf{R}$.

Solution

We have,

$y=x^{2} e^{-x}$ $\therefore \frac{d y}{d x}=2 x e^{-x}-x^{2} e^{-x}=x e^{-x}(2-x)$

Now, $\frac{d y}{d x}=0$.

$\Rightarrow x=0$ and $x=2$

The points $x=0$ and $x=2$ divide the real line into three disjoint intervals

i.e., $(-\infty, 0),(0,2)$, and $(2, \infty)$.

In intervals $(-\infty, 0)$ and $(2, \infty), f^{\prime}(x)<0$ as $e^{-x}$ is always positive.

$\therefore f$ is decreasing on $(-\infty, 0)$ and $(2, \infty)$.

In interval $(0,2), f^{\prime}(x)>0$.

$\therefore f$ is strictly increasing on $(0,2)$.

Hence, $f$ is strictly increasing in interval $(0,2)$.

The correct answer is D.

Solution

(i) The given function is $f(x)=(2 x-1)^{2}+3$.

It can be observed that $(2 x-1)^{2} \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=(2 x-1)^{2}+3 \geq 3$ for every $x \square \mathbf{R}$.

The minimum value of $f$ is attained when $2 x-1=0$.

$2 x-1=0$

$ x=\frac{1}{2} $

$\square$ Minimum value of $f=f(\frac{1}{2})=(2 \cdot \frac{1}{2}-1)^{2}+3=3$

Hence, function $f$ does not have a maximum value.

(ii) The given function is $f(x)=9 x^{2}+12 x+2=(3 x+2)^{2}-2$.

It can be observed that $(3 x+2)^{2} \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=(3 x+2)^{2}-2 \geq-2$ for every $x \square \mathbf{R}$.

The minimum value of $f$ is attained when $3 x+2=0$.

$3 x+2=0 \square^{x=\frac{-2}{3}}$

$\square$ Minimum value of $f=f(-\frac{2}{3})=(3(\frac{-2}{3})+2)^{2}-2=-2$

Hence, function $f$ does not have a maximum value.

(iii) The given function is $f(x)=-(x-1)^{2}+10$.

It can be observed that $(x-1)^{2} \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=-(x-1)^{2}+10 \leq 10$ for every $x \square \mathbf{R}$.

The maximum value of $f$ is attained when $(x-1)=0$.

$(x-1)=0 \square x=0$

$\square$ Maximum value of $f=f(1)=-(1-1)^{2}+10=10$

Hence, function $f$ does not have a minimum value.

(iv) The given function is $g(x)=x^{3}+1$.

Hence, function $g$ neither has a maximum value nor a minimum value.

Solution

(i) $f(x)=|x+2|-1$

We know that $|x+2| \geq 0$ for every $x \square \mathbf{R}$.

Therefore, $f(x)=|x+2|-1 \geq-1$ for every $x \square \mathbf{R}$.

The minimum value of $f$ is attained when $|x+2|=0$.

$|x+2|=0$

$\Rightarrow x=-2$

$\square$ Minimum value of $f=f(-2)=|-2+2|-1=-1$

Hence, function $f$ does not have a maximum value.

(ii) $g(x)=-|x+1|+3$

We know that $-|x+1| \leq 0$ for every $x \square \mathbf{R}$.

Therefore, $g(x)=-|x+1|+3 \leq 3$ for every $x \square \mathbf{R}$.

The maximum value of $g$ is attained when $|x+1|=0$.

$|x+1|=0$

$\Rightarrow x=-1$

$\square$ Maximum value of $g=g(-1)=-|-1+1|+3=3$

Hence, function $g$ does not have a minimum value. (iii) $h(x)=\sin 2 x+5$

We know that $-1 \leq \sin 2 x \leq 1$.

$\square-1+5 \leq \sin 2 x+5 \leq 1+5$

$\square 4 \leq \sin 2 x+5 \leq 6$

Hence, the maximum and minimum values of $h$ are 6 and 4 respectively.

(iv) $f(x)=|\sin 4 x+3|$

We know that $-1 \leq \sin 4 x \leq 1$.

$\square 2 \leq \sin 4 x+3 \leq 4$

$\square 2 \leq^{|\sin 4 x+3|} \leq 4$

Hence, the maximum and minimum values of $f$ are 4 and 2 respectively.

(v) $h(x)=x+1, x \square(-1,1)$

Here, if a point $x_0$ is closest to -1 , then we find $\frac{x_0}{2}+1<x_0+1$ for all $x_0 \square(-1,1)$.

Also, if $x_1$ is closest to 1 , then $x_1+1<\frac{x_1+1}{2}+1$ for all $x_1 \square(-1,1)$.

Hence, function $h(x)$ has neither maximum nor minimum value in $(-1,1)$.

Solution

(i) $f(x)=x^{2}$

$\therefore f^{\prime}(x)=2 x$

Now,

$f^{\prime}(x)=0 \Rightarrow x=0$

Thus, $x=0$ is the only critical point which could possibly be the point of local maxima or local minima of $f$.

We have $f^{\prime \prime}(0)=2$, which is positive.

Therefore, by second derivative test, $x=0$ is a point of local minima and local minimum value of $f$ at $x=0$ is $f(0)=0$.

(ii) $g(x)=x^{3}-3 x$

$\therefore g^{\prime}(x)=3 x^{2}-3$

Now,

$g^{\prime}(x)=0 \Rightarrow 3 x^{2}=3 \Rightarrow x= \pm 1$

$g^{\prime}(x)=6 x$

$g^{\prime}(1)=6>0$

$g^{\prime}(-1)=-6<0$

By second derivative test, $x=1$ is a point of local minima and local minimum value of $g$ at $x=1$ is $g(1)=1^{3}-3=1-3=-2$. However, $x=-1$ is a point of local maxima and local maximum value of $g$ at $x=-1$ is $g(1)=(-1)^{3}-3(-1)=-1+3=2$.

(iii) $h(x)=\sin x+\cos x, 0<x<\frac{\pi}{2}$

$\therefore h^{\prime}(x)=\cos x-\sin x$

$h^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \in(0, \frac{\pi}{2})$

$h^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)$

$h^{\prime \prime}(\frac{\pi}{4})=-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$

Therefore, by second derivative test, $x=\frac{\pi}{4}$ is a point of local maxima and the local

maximum value of $h$ at $x=\frac{\pi}{4}$ is $h(\frac{\pi}{4})=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$.

(iv) $f(x)=\sin x-\cos x, 0<x<2 \pi$

$\therefore f^{\prime}(x)=\cos x+\sin x$

$f^{\prime}(x)=0 \Rightarrow \cos x=-\sin x \Rightarrow \tan x=-1 \Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \in(0,2 \pi)$

$f^{\prime \prime}(x)=-\sin x+\cos x$

$f^{\prime \prime}(\frac{3 \pi}{4})=-\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}>0$

$f^{\prime \prime}(\frac{7 \pi}{4})=-\sin \frac{7 \pi}{4}+\cos \frac{7 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$

Therefore, by second derivative test, $x=\frac{3 \pi}{4}$ is a point of local maxima and the local

maximum value of $f$ at $x=\frac{3 \pi}{4}$ is

$f(\frac{3 \pi}{4})=\sin \frac{3 \pi}{4}-\cos \frac{3 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$ However, $x=\frac{7 \pi}{4}$ is a point of local minima and

the local minimum value of $f$ at $x=\frac{7 \pi}{4}$ is $f(\frac{7 \pi}{4})=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}$.

(v) $f(x)=x^{3}-6 x^{2}+9 x+15$

$\therefore f^{\prime}(x)=3 x^{2}-12 x+9$

$f^{\prime}(x)=0 \Rightarrow 3(x^{2}-4 x+3)=0$

$\Rightarrow 3(x-1)(x-3)=0$

$\Rightarrow x=1,3$

Now, $f^{\prime \prime}$

$(x)=6 x-12=6(x-2)$

$f^{\prime \prime}(1)=6(1-2)=-6<0$

$f^{\prime \prime}(3)=6(3-2)=6>0$

Therefore, by second derivative test, $x=1$ is a point of local maxima and the local maximum value of $f$ at $x=1$ is $f(1)=1-6+9+15=19$. However, $x=3$ is a point of local minima and the local minimum value of $f$ at $x=3$ is $f(3)=27-54+27+15=$ 15.

(vi) $g(x)=\frac{x}{2}+\frac{2}{x}, x>0$

$\therefore g^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$

Now,

$g^{\prime}(x)=0$ gives $\frac{2}{x^{2}}=\frac{1}{2} \Rightarrow x^{2}=4 \Rightarrow x= \pm 2$

Since $x>0$, we take $x=2$.

Now,

$g^{\prime \prime}(x)=\frac{4}{x^{3}}$

$g^{\prime \prime}(2)=\frac{4}{2^{3}}=\frac{1}{2}>0$

Therefore, by second derivative test, $x=2$ is a point of local minima and the local

minimum value of $g$ at $x=2$ is $g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2$.

(vii) $g(x)=\frac{1}{x^{2}+2}$

$\therefore g^{\prime}(x)=\frac{-(2 x)}{(x^{2}+2)^{2}}$

$g^{\prime}(x)=0 \Rightarrow \frac{-2 x}{(x^{2}+2)^{2}}=0 \Rightarrow x=0$

Now, for values close to $x=0$ and to the left of $0, g^{\prime}(x)>0$. Also, for values close to $x=$ 0 and to the right of $0, g^{\prime}(x)<0$.

Therefore, by first derivative test, $x=0$ is a point of local maxima and the local maximum value of $g(0)$ is $\frac{1}{0+2}=\frac{1}{2}$.

(viii) $f(x)=x \sqrt{1-x}, x>0$

$\therefore f^{\prime}(x)=\sqrt{1-x}+x \cdot \frac{1}{2 \sqrt{1-x}}(-1)=\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}}$

$=\frac{2(1-x)-x}{2 \sqrt{1-x}}=\frac{2-3 x}{2 \sqrt{1-x}}$

$f^{\prime}(x)=0 \Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0 \Rightarrow 2-3 x=0 \Rightarrow x=\frac{2}{3}$

$f^{\prime \prime}(x)=\frac{1}{2}[\frac{\sqrt{1-x}(-3)-(2-3 x)(\frac{-1}{2 \sqrt{1-x}})}{1-x}]$

$=\frac{\sqrt{1-x}(-3)+(2-3 x)(\frac{1}{2 \sqrt{1-x}})}{2(1-x)}$

$=\frac{-6(1-x)+(2-3 x)}{4(1-x)^{\frac{3}{2}}}$

$=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}$

$f^{\prime \prime}(\frac{2}{3})=\frac{3(\frac{2}{3})-4}{4(1-\frac{2}{3})^{\frac{3}{2}}}=\frac{2-4}{4(\frac{1}{3})^{\frac{3}{2}}}=\frac{-1}{2(\frac{1}{3})^{\frac{3}{2}}}<0$

Therefore, by second derivative test, $x=\frac{2}{3}$ is a point of local maxima and the local

maximum value of $f$ at $x=\frac{2}{3}$ is $f(\frac{2}{3})=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9}$.

Solution

i. We have,

$f(x)=e^{x}$

$\therefore f^{\prime}(x)=e^{x}$

Now, if $f^{\prime}(x)=0$, then $e^{x}=0$. But, the exponential function can never assume 0 for any value of $x$.

Therefore, there does not exist $c \square \mathbf{R}$ such that $f^{\prime}(c)=0$.

Hence, function $f$ does not have maxima or minima.

ii. We have,

$g(x)=\log x$

$\therefore g^{\prime}(x)=\frac{1}{x}$

Since $\log x$ is defined for a positive number $x, g^{\prime}(x)>0$ for any $x$.

Therefore, there does not exist $c \square \mathbf{R}$ such that $g^{\prime}(c)=0$.

Hence, function $g$ does not have maxima or minima.

iii. We have,

$h(x)=x^{3}+x^{2}+x+1$

$\therefore h^{\prime}(x)=3 x^{2}+2 x+1$

Now,

$h(x)=0 \square 3 x^{2}+2 x+1=0 \square x=\frac{-2 \pm 2 \sqrt{2} i}{6}=\frac{-1 \pm \sqrt{2} i}{3} \notin \mathbf{R}$

Therefore, there does not exist $c \square \mathbf{R}$ such that $h^{\prime}(c)=0$.

Hence, function $h$ does not have maxima or minima.

Solution

(i) The given function is $f(x)=x^{3}$.

$\therefore f^{\prime}(x)=3 x^{2}$

Now,

$f^{\prime}(x)=0 \Rightarrow x=0$

Then, we evaluate the value of $f$ at critical point $x=0$ and at end points of the interval $[-2,2]$.

$f(0)=0$

$f(-2)=(-2)^{3}=-8$

$f(2)=(2)^{3}=8$

Hence, we can conclude that the absolute maximum value of $f$ on $[-2,2]$ is 8 occurring at $x=2$. Also, the absolute minimum value of $f$ on $[-2,2]$ is -8 occurring at $x=-2$.

(ii) The given function is $f(x)=\sin x+\cos x$.

$\therefore f^{\prime}(x)=\cos x-\sin x$

Now,

$f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}$

Then, we evaluate the value of $f$ at critical point $x=\frac{\pi}{4}$ and at the end points of the interval $[0, \pi]$.

Solution

The profit function is given as $p(x)=41-24 x-18 x^{2}$.

$\therefore p^{\prime}(x)=-24-36 x$

$p^{\prime \prime}(x)=-36$

Now,

$p^{\prime}(x)=0 \Rightarrow x=\frac{-24}{36}=-\frac{2}{3}$

Also,

$p^{\prime \prime}(\frac{-2}{3})=-36<0$

By second derivative test, $x=-\frac{2}{3}$ is the point of local maxima of $p$.

$\therefore$ Maximum profit $=p(-\frac{2}{3})$

$ \begin{aligned} & =41-24(-\frac{2}{3})-18(-\frac{2}{3})^{2} \\ & =41+16-8 \\ & =49 \end{aligned} $

Hence, the maximum profit that the company can make is 49 units.

Solution

$f(x)=3 x^4-8 x^3+12 x^2-48 x+25$

$f^{\prime}(x)=0$ for finding critical points

$f^{\prime}(x)=12 x^3-24 x^2+24 x-48$

$12 x^3-24 x^2+24 x-48=0$

$12 x^2(x-2)+24(x-2)=0$

$\left(12 x^2+24\right)(x-2)=0$

Since $12 x^2+24 \neq 0$ and $x-2=0 \Longrightarrow x=2$

We need to check at $x=0,2,3$ $f(0)=25$

$f(2)=-39$

$f(3)=16$

$\therefore$ Maximum of $f(x)$ at $x=0$ is 25

Minimum of $f(x)$ at $x=2$ is -39

Solution

Let $f(x)=\sin 2 x$.

$\therefore f^{\prime}(x)=2 \cos 2 x$

Now,

$ \begin{aligned} & f^{\prime}(x)=0 \Rightarrow \cos 2 x=0 \\ & \Rightarrow 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \end{aligned} $

Then, we evaluate the values of $f$ at critical points $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ and at the end points of the interval $[0,2 \pi]$.

$f(\frac{\pi}{4})=\sin \frac{\pi}{2}=1, f(\frac{3 \pi}{4})=\sin \frac{3 \pi}{2}=-1$

$f(\frac{5 \pi}{4})=\sin \frac{5 \pi}{2}=1, f(\frac{7 \pi}{4})=\sin \frac{7 \pi}{2}=-1$

$f(0)=\sin 0=0, f(2 \pi)=\sin 2 \pi=0$

Hence, we can conclude that the absolute maximum value of $f$ on $[0,2 \pi]$ is occurring

at $x=\frac{\pi}{4}$ and $\quad x=\frac{5 \pi}{4}$.

Solution

Let $f(x)=\sin x+\cos x$.

$\therefore f^{\prime}(x)=\cos x-\sin x$

$f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4} \ldots$,

$f^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)$

Now, $f^{\prime \prime}(x)$ will be negative when $(\sin x+\cos x)$ is positive i.e., when $\sin x$ and $\cos x$ are both positive. Also, we know that $\sin x$ and $\cos x$ both are positive in the first

quadrant. Then, $f^{\prime \prime}(x)$ will be negative when $x \in(0, \frac{\pi}{2})$.

Thus, we consider $x=\frac{\pi}{4}$.

$f^{\prime \prime}(\frac{\pi}{4})=-(\sin \frac{\pi}{4}+\cos \frac{\pi}{4})=-(\frac{2}{\sqrt{2}})=-\sqrt{2}<0$ $\square$ By second derivative test, $f$ will be the maximum at $x=\frac{\pi}{4}$ and the maximum value of $f$ is $f(\frac{\pi}{4})=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$.

Solution

Let $f(x)=2 x^{3}-24 x+107$.

$\therefore f^{\prime}(x)=6 x^{2}-24=6(x^{2}-4)$

Now,

$f^{\prime}(x)=0 \Rightarrow 6(x^{2}-4)=0 \Rightarrow x^{2}=4 \Rightarrow x= \pm 2$

We first consider the interval $[1,3]$.

Then, we evaluate the value of $f$ at the critical point $x=2 \square[1,3]$ and at the end points of the interval $[1,3]$.

$f(2)=2(8)-24(2)+107=16-48+107=75$

$f(1)=2(1)-24(1)+107=2-24+107=85$

$f(3)=2(27)-24(3)+107=54-72+107=89$

Hence, the absolute maximum value of $f(x)$ in the interval $[1,3]$ is 89 occurring at $x=$ 3.

Next, we consider the interval $[-3,-1]$.

Evaluate the value of $f$ at the critical point $x=-2 \square[-3,-1]$ and at the end points of the interval $[1,3]$.

$f(-3)=2(-27)-24(-3)+107=-54+72+107=125$

$f(-1)=2(-1)-24(-1)+107=-2+24+107=129$

$f(-2)=2(-8)-24(-2)+107=-16+48+107=139$

Hence, the absolute maximum value of $f(x)$ in the interval $[-3,-1]$ is 139 occurring at $x$ $=-2$.

Solution

Let $f(x)=x^{4}-62 x^{2}+a x+9$.

$\therefore f^{\prime}(x)=4 x^{3}-124 x+a$

It is given that function $f$ attains its maximum value on the interval $[0,2]$ at $x=1$.

$\therefore f^{\prime}(1)=0$

$\Rightarrow 4-124+a=0$

$\Rightarrow a=120$

Hence, the value of $a$ is 120 .

Solution

Let $f(x)=x+\sin 2 x$.

$\therefore f^{\prime}(x)=1+2 \cos 2 x$

Now, $f^{\prime}(x)=0 \Rightarrow \cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos (\pi-\frac{\pi}{3})=\cos \frac{2 \pi}{3}$

$2 x=2 \pi \pm \frac{2 \pi}{3} \quad n \in \mathbf{Z}$

$\Rightarrow x=n \pi \pm \frac{\pi}{3}, n \in \mathbf{Z}$

$\Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \in[0,2 \pi]$

Then, we evaluate the value of $f$ at critical points $x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}$ and at the end points of the interval $[0,2 \pi]$. $f(\frac{\pi}{3})=\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}$

$f(\frac{2 \pi}{3})=\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$

$f(\frac{4 \pi}{3})=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}$

$f(\frac{5 \pi}{3})=\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}$

$f(0)=0+\sin 0=0$

$f(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi$

Hence, we can conclude that the absolute maximum value of $f(x)$ in the interval $[0,2 \pi]$ is $2 \pi$ occurring at $x=2 \pi$ and the absolute minimum value of $f(x)$ in the interval $[0,2 \pi]$ is 0 occurring at $x=0$.

Solution

Let one number be $x$. Then, the other number is $(24-x)$.

Let $P(x)$ denote the product of the two numbers. Thus, we have:

$ \begin{aligned} & P(x)=x(24-x)=24 x-x^{2} \\ & \therefore P^{\prime}(x)=24-2 x \\ & P^{\prime \prime}(x)=-2 \end{aligned} $

Now,

$P^{\prime}(x)=0 \Rightarrow x=12$

Also,

$P^{\prime \prime}(12)=-2<0$

$\square$ By second derivative test, $x=12$ is the point of local maxima of $P$. Hence, the product of the numbers is the maximum when the numbers are 12 and $24-12=12$.

Solution

The two numbers are $x$ and $y$ such that $x+y=60$.

$\square y=60-x$

Let $f(x)=x y^{3}$.

$\Rightarrow f(x)=x(60-x)^{3}$

$\therefore f^{\prime}(x)=(60-x)^{3}-3 x(60-x)^{2}$

$=(60-x)^{2}[60-x-3 x]$

$=(60-x)^{2}(60-4 x)$

And, $f^{\prime \prime}(x)=-2(60-x)(60-4 x)-4(60-x)^{2}$

$ \begin{aligned} & =-2(60-x)[60-4 x+2(60-x)] \\ & =-2(60-x)(180-6 x) \\ & =-12(60-x)(30-x) \end{aligned} $

Now, $f^{\prime}(x)=0 \Rightarrow x=60$ or $x=15$

When $x=60, f^{\prime \prime}(x)=0$.

When $x=15, f^{\prime \prime}(x)=-12(60-15)(30-15)=-12 \times 45 \times 15<0$.

$\square$ By second derivative test, $x=15$ is a point of local maxima of $f$. Thus, function $x y^{3}$ is maximum when $x=15$ and $y=60-15=45$.

Hence, the required numbers are 15 and 45 .

Solution

Let one number be $x$. Then, the other number is $y=(35-x)$.

Let $P(x)=x^{2} y^{5}$. Then, we have:

$ \begin{aligned} & P(x)=x^{2}(35-x)^{5} \\ & \begin{aligned} \therefore P^{\prime}(x) & =2 x(35-x)^{5}-5 x^{2}(35-x)^{4} \\ & =x(35-x)^{4}[2(35-x)-5 x] \\ & =x(35-x)^{4}(70-7 x) \\ & =7 x(35-x)^{4}(10-x) \end{aligned} \end{aligned} $

And, $P^{\prime \prime}(x)=7(35-x)^{4}(10-x)+7 x[-(35-x)^{4}-4(35-x)^{3}(10-x)]$

$ \begin{aligned} & =7(35-x)^{4}(10-x)-7 x(35-x)^{4}-28 x(35-x)^{3}(10-x) \\ & =7(35-x)^{3}[(35-x)(10-x)-x(35-x)-4 x(10-x)] \\ & =7(35-x)^{3}[350-45 x+x^{2}-35 x+x^{2}-40 x+4 x^{2}] \\ & =7(35-x)^{3}(6 x^{2}-120 x+350) \end{aligned} $

Now, $P^{\prime}(x)=0 \Rightarrow x=0, x=35, x=10$

When $x=35,{ }^{\prime}(x)=f(x)=0$ and $y=35-35=0$. This will make the product $x^{2} y^{5}$ equal to 0 .

When $x=0, y=35-0=35$ and the product $x^{2} y^{2}$ will be 0 .

$\square x=0$ and $x=35$ cannot be the possible values of $x$.

When $x=10$, we have:

$ \begin{aligned} P^{\prime \prime}(x) & =7(35-10)^{3}(6 \times 100-120 \times 10+350) \\ & =7(25)^{3}(-250)<0 \end{aligned} $

$\square$ By second derivative test, $P(x)$ will be the maximum when $x=10$ and $y=35-10=$ 25.

Hence, the required numbers are 10 and 25 .

Solution

Let one number be $x$. Then, the other number is $(16-x)$.

Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,

$S(x)=x^{3}+(16-x)^{3}$

$\therefore S^{\prime}(x)=3 x^{2}-3(16-x)^{2}, S^{\prime \prime}(x)=6 x+6(16-x)$

Now, $S^{\prime}(x)=0 \Rightarrow 3 x^{2}-3(16-x)^{2}=0$

$\Rightarrow x^{2}-(16-x)^{2}=0$

$\Rightarrow x^{2}-256-x^{2}+32 x=0$

$\Rightarrow x=\frac{256}{32}=8$

Now, $S^{\prime \prime}(8)=6(8)+6(16-8)=48+48=96>0$

$\square$ By second derivative test, $x=8$ is the point of local minima of $S$.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16-8=8$.

Solution

Let the side of the square to be cut off be $x cm$. Then, the length and the breadth of the box will be $(18-2 x) cm$ each and the height of the box is $x cm$.

Therefore, the volume $V(x)$ of the box is given by,

$V(x)=x(18-2 x)^{2}$ $\therefore V^{\prime}(x)=(18-2 x)^{2}-4 x(18-2 x)$

$=(18-2 x)[18-2 x-4 x]$

$=(18-2 x)(18-6 x)$

$=6 \times 2(9-x)(3-x)$

$=12(9-x)(3-x)$

And, $V^{\prime \prime}(x)=12[-(9-x)-(3-x)]$

$=-12(9-x+3-x)$

$=-12(12-2 x)$

$=-24(6-x)$

Now, $V^{\prime}(x)=0 \Rightarrow x=9$ or $x=3$

If $x=9$, then the length and the breadth will become 0 .

$\therefore x \neq 9$.

$\Rightarrow x=3$.

Now, $V^{\prime \prime}(3)=-24(6-3)=-72<0$

$\therefore$ By second derivative test, $x=3$ is the point of maxima of $V$.

Hence, if we remove a square of side $3 cm$ from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Solution

Let the side of the square to be cut off be $x cm$. Then, the height of the box is $x$, the length is $45-2 x$, and the breadth is $24-2 x$.

Therefore, the volume $V(x)$ of the box is given by,

$ \begin{aligned} V(x) & =x(45-2 x)(24-2 x) \\ & =x(1080-90 x-48 x+4 x^{2}) \\ & =4 x^{3}-138 x^{2}+1080 x \end{aligned} $

$\therefore V^{\prime}(x)=12 x^{2}-276 x+1080$

$ =12(x^{2}-23 x+90) $

$ =12(x-18)(x-5) $

$V^{\prime \prime}(x)=24 x-276=12(2 x-23)$

Now, ${ }^{V^{\prime}}(x)=0 \Rightarrow x=18$ and $x=5$

It is not possible to cut off a square of side $18 cm$ from each corner of the rectangular sheet. Thus, $x$ cannot be equal to 18 .

$\square x=5$

Now, $V^{\prime \prime}(5)=12(10-23)=12(-13)=-156<0$

$\therefore$ By second derivative test, $x=5$ is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is $5 cm$.

Solution

Let a rectangle of length / and breadth $b$ be inscribed in the given circle of radius $a$. Then, the diagonal passes through the centre and is of length $2 a cm$.

Now, by applying the Pythagoras theorem, we have:

$ \begin{aligned} & (2 a)^{2}=l^{2}+b^{2} \\ & \Rightarrow b^{2}=4 a^{2}-l^{2} \\ & \Rightarrow b=\sqrt{4 a^{2}-l^{2}} \end{aligned} $

$\square$ Area of the rectangle, $A=l \sqrt{4 a^{2}-l^{2}}$

$\therefore \frac{d A}{d l}=\sqrt{4 a^{2}-l^{2}}+l \frac{1}{2 \sqrt{4 a^{2}-l^{2}}}(-2 l)=\sqrt{4 a^{2}-l^{2}}-\frac{l^{2}}{\sqrt{4 a^{2}-l^{2}}}$

$ =\frac{4 a^{2}-2 l^{2}}{\sqrt{4 a^{2}-l^{2}}} $

$\frac{d^{2} A}{d l^{2}}=\frac{\sqrt{4 a^{2}-l^{2}}(-4 l)-(4 a^{2}-2 l^{2}) \frac{(-2 l)}{2 \sqrt{4 a^{2}-l^{2}}}}{(4 a^{2}-l^{2})}$

$ =\frac{(4 a^{2}-l^{2})(-4 l)+l(4 a^{2}-2 l^{2})}{(4 a^{2}-l^{2})^{\frac{3}{2}}} $

$ =\frac{-12 a^{2} l+2 l^{3}}{(4 a^{2}-l^{2})^{\frac{3}{2}}}=\frac{-2 l(6 a^{2}-l^{2})}{(4 a^{2}-l^{2})^{\frac{3}{2}}} $

Now, $\frac{d A}{d l}=0$ gives $4 a^{2}=2 l^{2} \Rightarrow l=\sqrt{2} a$

$ \Rightarrow b=\sqrt{4 a^{2}-2 a^{2}}=\sqrt{2 a^{2}}=\sqrt{2} a $

Now, when $l=\sqrt{2} a$,

$\frac{d^{2} A}{d l^{2}}=\frac{-2(\sqrt{2} a)(6 a^{2}-2 a^{2})}{2 \sqrt{2} a^{3}}=\frac{-8 \sqrt{2} a^{3}}{2 \sqrt{2} a^{3}}=-4<0$

$\therefore$ By the second derivative test, when $l=\sqrt{2} a$, then the area of the rectangle is the maximum.

Since $l=b=\sqrt{2} a$, the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Solution

Let $r$ and $h$ be the radius and height of the cylinder respectively.

Then, the surface area $(S)$ of the cylinder is given by,

$ \begin{aligned} & S=2 \pi r^{2}+2 \pi r h \\ & \Rightarrow h=\frac{S-2 \pi r^{2}}{2 \pi r} \\ &=\frac{S}{2 \pi}(\frac{1}{r})-r \end{aligned} $

Let $V$ be the volume of the cylinder. Then,

$V=\pi r^{2} h=\pi r^{2}[\frac{S}{2 \pi}(\frac{1}{r})-r]=\frac{S r}{2}-\pi r^{3}$

Then, $\frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}, \frac{d^{2} V}{d r^{2}}=-6 \pi r$

Now, $\frac{d V}{d r}=0 \Rightarrow \frac{S}{2}=3 \pi r^{2} \Rightarrow r^{2}=\frac{S}{6 \pi}$

When $r^{2}=\frac{S}{6 \pi}$, then $\frac{d^{2} V}{d r^{2}}=-6 \pi(\sqrt{\frac{S}{6 \pi}})<0$.

By second derivative test, the volume is the maximum when $r^{2}=\frac{S}{6 \pi}$.

Now, when $r^{2}=\frac{S}{6 \pi}$, then $h=\frac{6 \pi r^{2}}{2 \pi}(\frac{1}{r})-r=3 r-r=2 r$.

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

Solution

Let $r$ and $h$ be the radius and height of the cylinder respectively.

Then, volume $(V)$ of the cylinder is given by,

$V=\pi r^{2} h=100 \quad$ (given)

$\therefore h=\frac{100}{\pi r^{2}}$

Surface area $(S$ ) of the cylinder is given by,

$S=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+\frac{200}{r}$

$\therefore \frac{d S}{d r}=4 \pi r-\frac{200}{r^{2}}, \frac{d^{2} S}{d r^{2}}=4 \pi+\frac{400}{r^{3}}$

$\frac{d S}{d r}=0 \Rightarrow 4 \pi r=\frac{200}{r^{2}}$

$\Rightarrow r^{3}=\frac{200}{4 \pi}=\frac{50}{\pi}$

$\Rightarrow r=(\frac{50}{\pi})^{\frac{1}{3}}$

Now, it is observed that when $r=(\frac{50}{\pi})^{\frac{1}{3}}, \frac{d^{2} S}{d r^{2}}>0$.

$\square$ By second derivative test, the surface area is the minimum when the radius of the

cylinder is $(\frac{50}{\pi})^{\frac{1}{3}} cm$.

When $r=(\frac{50}{\pi})^{\frac{1}{3}}, h=\frac{100}{\pi(\frac{50}{\pi})^{\frac{2}{3}}}=\frac{2 \times 50}{(50)^{\frac{2}{3}}()^{1-\frac{2}{3}}}=2(\frac{50}{\pi})^{\frac{1}{3}} cm$.

Hence, the required dimensions of the can which has the minimum surface area is given

by radius $=(\frac{50}{\pi})^{\frac{1}{3}} cm$ and height $=2(\frac{50}{\pi})^{\frac{1}{3}} cm$.

Solution

Let a piece of length / be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length $(28-I) m$.

Now, side of square $=\frac{l}{4}$.

Let $r$ be the radius of the circle. Then,

$ 2 \pi r=28-l \Rightarrow r=\frac{1}{2 \pi}(28-l) \text{. } $

The combined areas of the square and the circle $(A)$ is given by,

$A=(\text{ side of the square })^{2}+r^{2}$

$=\frac{l^{2}}{16}+\pi[\frac{1}{2 \pi}(28-l)]^{2}$

$=\frac{l^{2}}{16}+\frac{1}{4 \pi}(28-l)^{2}$

$\therefore \frac{d A}{d l}=\frac{2 l}{16}+\frac{2}{4 \pi}(28-l)(-1)=\frac{l}{8}-\frac{1}{2 \pi}(28-l)$

$\frac{d^{2} A}{d l^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0$

Now, $\frac{d A}{d l}=0 \Rightarrow \frac{l}{8}-\frac{1}{2 \pi}(28-l)=0$

$\Rightarrow \frac{\pi l-4(28-l)}{8 \pi}=0$

$\Rightarrow(\pi+4) l-112=0$

$\Rightarrow l=\frac{112}{\pi+4}$

Thus, when $l=\frac{112}{\pi+4}, \frac{d^{2} A}{d l^{2}}>0$.

$\therefore$ By second derivative test, the area $(A)$ is the minimum when $l=\frac{112}{\pi+4}$.

Hence, the combined area is the minimum when the length of the wire in making the

square is $\frac{112}{\pi+4} cm$ while the length of the wire in making the circle

is $28-\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4} cm$.

Solution

Let $r$ and $h$ be the radius and height of the cone respectively inscribed in a sphere of radius $R$.

Let $V$ be the volume of the cone.

Then, $V=\frac{1}{3} \pi r^{2} h$

Height of the cone is given by,

$h=R+AB=R+\sqrt{R^{2}-r^{2}} \quad$ [ $ABC$ is a right triangle]

$ \begin{aligned} & \therefore V=\frac{1}{3} \pi r^{2}(R+\sqrt{R^{2}-r^{2}}) \\ & =\frac{1}{3} \pi r^{2} R+\frac{1}{3} \pi r^{2} \sqrt{R^{2}-r^{2}} \\ & \therefore \frac{d V}{d r}=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}+\frac{1}{3} \pi r^{2} \cdot \frac{(-2 r)}{2 \sqrt{R^{2}-r^{2}}} \\ & =\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}-\frac{1}{3} \pi \frac{r^{3}}{\sqrt{R^{2}-r^{2}}} \\ & =\frac{2}{3} \pi r R+\frac{2 \pi r(R^{2}-r^{2})-\pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \\ & =\frac{2}{3} \pi r R+\frac{2 \pi r R^{2}-3 \pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \\ & \frac{d^{2} V}{d r^{2}}=\frac{2 \pi R}{3}+\frac{3 \sqrt{R^{2}-r^{2}}(2 \pi R^{2}-9 \pi r^{2})-(2 \pi r R^{2}-3 \pi r^{3}) \cdot \frac{(-2 r)}{6 \sqrt{R^{2}-r^{2}}}}{9(R^{2}-r^{2})} \\ & =\frac{2}{3} \pi R+\frac{9(R^{2}-r^{2})(2 \pi R^{2}-9 \pi r^{2})+2 \pi r^{2} R^{2}+3 \pi r^{4}}{27(R^{2}-r^{2})^{\frac{3}{2}}} \end{aligned} $

Now, $\frac{d V}{d r}=0 \Rightarrow \frac{2}{3} \quad r R=\frac{3 \pi r^{3}-2 \pi r R^{2}}{3 \sqrt{R^{2}-r^{2}}}$

$\Rightarrow 2 R=\frac{3 r^{2}-2 R^{2}}{\sqrt{R^{2}-r^{2}}} \Rightarrow 2 R \sqrt{R^{2}-r^{2}}=3 r^{2}-2 R^{2}$

$\Rightarrow 4 R^{2}(R^{2}-r^{2})=(3 r^{2}-2 R^{2})^{2}$

$\Rightarrow 4 R^{4}-4 R^{2} r^{2}=9 r^{4}+4 R^{4}-12 r^{2} R^{2}$

$\Rightarrow 9 r^{4}=8 R^{2} r^{2}$

$\Rightarrow r^{2}=\frac{8}{9} R^{2}$

When $r^{2}=\frac{8}{9} R^{2}$, then $\frac{d^{2} V}{d r^{2}}<0$.

By second derivative test, the volume of the cone is the maximum when $r^{2}=\frac{8}{9} R^{2}$.

When $r^{2}=\frac{8}{9} R^{2}, h=R+\sqrt{R^{2}-\frac{8}{9} R^{2}}=R+\sqrt{\frac{1}{9} R^{2}}=R+\frac{R}{3}=\frac{4}{3} R$.

Therefore,

$=\frac{1}{3} \pi(\frac{8}{9} R^{2})(\frac{4}{3} R)$

$=\frac{8}{27}(\frac{4}{3} \pi R^{3})$

$=\frac{8}{27} \times($ Volume of the sphere $)$

Hence, the volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$ the volume of the sphere.

Solution

Let $r$ and $h$ be the radius and the height (altitude) of the cone respectively.

Then, the volume $(V)$ of the cone is given as:

$V=\frac{1}{3} \pi r^{2} h \Rightarrow h=\frac{3 V}{r^{2}}$

The surface area $(S)$ of the cone is given by,

$S=\pi r l$ (where $/$ is the slant height)

$=\pi r \sqrt{r^{2}+h^{2}}$

$=\pi r \sqrt{r^{2}+\frac{9 \hbar^{2}}{\pi^{2} r^{4}}} \stackrel{\pi}{=} \frac{r \sqrt{9^{2} r^{6}+V^{2}}}{\pi r^{2}}$

$=\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{2}}$

$ \begin{aligned} \therefore \frac{d S}{d r} & =\frac{r \cdot \frac{6 \pi^{2} r^{5}}{2 \sqrt{{ }^{2} r^{6} 9 V^{2}}}-\sqrt{\pi^{2} r^{6}+9 V^{2}}}{r^{2}} \\ & =\frac{3 \pi^{2} r^{6}-\pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \\ & =\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \\ & =\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \end{aligned} $

Now, $\frac{d S}{d r}=0 \Rightarrow 2 \pi^{2} r^{6}=9 V^{2} \Rightarrow r^{6}=\frac{9 V^{2}}{2 \pi^{2}}$

Thus, it can be easily verified that when $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, \frac{d^{2} S}{d r^{2}}>0$.

$\square$ By second derivative test, the surface area of the cone is the least when $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}$.

When $r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, h=\frac{3 V}{\pi r^{2}}=\frac{3}{\pi r^{2}}(\frac{2 \pi^{2} r^{6}}{9})^{\frac{1}{2}}=\frac{3}{\pi r^{2}} \cdot \frac{\sqrt{2} \pi r^{3}}{3}=\sqrt{2} r$.

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to $\sqrt{2}$ times the radius of the base.

Solution

Let $\theta$ be the semi-vertical angle of the cone.

It is clear that

$ \theta \in[0, \frac{\pi}{2}] $

Let $r, h$, and $/$ be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant.

Now, $r=I \sin \theta$ and $h=I \cos \theta$

The volume $(V)$ of the cone is given by,

$ \begin{aligned} & V=\frac{1}{3} \pi r^{2} h \\ & =\frac{1}{3} \pi(l^{2} \sin ^{2} \theta)(l \cos \theta) \\ & =\frac{1}{3} \pi l^{3} \sin ^{2} \theta \cos \theta \\ & \begin{aligned} \therefore \frac{d V}{d \theta} & =\frac{l^{3} \pi}{3}[\sin ^{2} \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)] \\ & =\frac{l^{3} \pi}{3}[-\sin ^{3}+2 \sin \theta \cos ^{2} \theta] \end{aligned} \\ & \begin{aligned} \frac{d^{2} V}{d \theta^{2}} & =\frac{l^{3} \pi}{3}[-3 \sin ^{2} \theta \cos \theta+2 \cos ^{3} \theta-4 \sin ^{2} \theta \cos \theta] \\ & =\frac{l^{3} \pi}{3}[2 \cos ^{3} \theta-7 \sin ^{2} \theta \cos \theta] \end{aligned} \end{aligned} $

Now, $\frac{d V}{d \theta}=0$

$\Rightarrow \sin ^{3} \theta=2 \sin \theta \cos ^{2} \theta$

$\Rightarrow \tan ^{2} \theta=2$

$\Rightarrow \tan \theta=\sqrt{2}$

$\Rightarrow \theta=\tan ^{-1} \sqrt{2}$

Now, when $\theta=\tan ^{-1} \sqrt{2}$, then $\tan ^{2} \theta=2$ or $\sin ^{2} \theta=2 \cos ^{2} \theta$.

Then, we have:

$\frac{d^{2} V}{d \theta^{2}}=\frac{l^{3} \pi}{3}[2 \cos ^{3} \theta-14 \cos ^{3} \theta]=-4 \pi l^{3} \cos ^{3} \theta<0$ for $\theta \in[0, \frac{\pi}{2}]$

$\square$ By second derivative test, the volume $(V)$ is the maximum when $\theta=\tan ^{-1} \sqrt{2}$.

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is $\tan ^{-1} \sqrt{2}$.

Solution

Let $\mathrm{r}, \mathrm{h}, \mathrm{l}$ be the radius, height and slant height of the right circular cone respectively. Let $S$ be the given surface area of the cone. We have, $l^2=\mathrm{r}^2+\mathrm{h}^2 \ldots .(1)$

$\begin{aligned} & S=\pi r l+\pi r^2 \ & S-\pi r^2=\pi r l \ & \Rightarrow l=\frac{S-\pi r^2}{\pi r} \ldots . . \end{aligned}$

$\begin{aligned} & \mathrm{V}=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\ & \Rightarrow \mathrm{V}=\frac{1}{3} \pi \mathrm{r}^2 \sqrt{1^2-\mathrm{r}^2}(\mathrm{by}(1)) \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \pi^2 \mathrm{r}^4\left(\mathrm{l}^2-\mathrm{r}^2\right) \end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{V}^2=\frac{1}{9} \pi^2 \mathrm{r}^4\left[\left(\frac{\mathrm{S}-\pi \mathrm{r}^2}{\pi \mathrm{r}}\right)^2-\mathrm{r}^2\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \pi^2 \mathrm{r}^4\left[\frac{\left(\mathrm{S}-\pi \mathrm{r}^2\right)^2-\pi^2 \mathrm{r}^4}{\pi^2 \mathrm{r}^2}\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \mathrm{r}^2\left[\left(\mathrm{~S}-\pi \mathrm{r}^2\right)^2-\pi^2 \mathrm{r}^4\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9} \mathrm{r}^2\left[\mathrm{~S}^2-2 \pi \mathrm{Sr}^2+\pi^2 \mathrm{r}^4-\pi^2 \mathrm{r}^4\right] \\ & \Rightarrow \mathrm{V}^2=\frac{1}{9}\left(\mathrm{r}^2 \mathrm{~S}^2-2 \pi \mathrm{Sr}^4\right) \\ \end{aligned}$

$2 \mathrm{~V} \frac{\mathrm{dV}}{\mathrm{dr}}=\frac{\mathrm{S}^2}{9} 2 \mathrm{r}-\frac{2 \pi \mathrm{S}}{9} 4 \mathrm{r}^3$

$2 \mathrm{~V} \frac{\mathrm{dV}}{\mathrm{dr}}=\frac{2 \mathrm{rS}}{9}\left(\mathrm{~S}-4 \pi \mathrm{r}^2\right)$

For maximum volume, $\frac{\mathrm{dV}}{\mathrm{dr}}=0$

$\begin{aligned} & \Rightarrow \frac{2 \mathrm{rS}}{9}\left(\mathrm{~S}-4 \pi \mathrm{r}^2\right)=0 \\ & \Rightarrow \mathrm{r}=\mathrm{o} \text { or } \mathrm{S}-4 \pi \mathrm{r}^2=0 \end{aligned}$

Since, $r$ cannot be $o$

$\begin{aligned} & \Rightarrow \mathrm{S}=4 \pi \mathrm{r}^2 \ & \Rightarrow \mathrm{r}^2=\frac{\mathrm{S}}{4 \pi} \ & \Rightarrow \mathrm{r}^2=\frac{\pi \mathrm{rl}+\pi \mathrm{r}^2}{4 \pi} \end{aligned}$

$\begin{aligned} & \Rightarrow 4 \pi r^2=\pi r l+\pi r^2 \ & \Rightarrow 3 \pi r^2=\pi r l \ & \Rightarrow l=3 r \end{aligned}$

Let $\alpha$ be the semi-vertical angle.

$\begin{aligned} & \sin \alpha=\frac{r}{l} \\ & \sin \alpha=\frac{r}{3 r} \\ & \Rightarrow \alpha=\sin ^{-1}\left(\frac{1}{3}\right) \end{aligned}$

Solution

The given curve is $x^{2}=2 y$.

For each value of $x$, the position of the point will be $(x, \frac{x^{2}}{2})$.

The distance $d(x)$ between the points $(x, \frac{x^{2}}{2})$ and $(0,5)$ is given by,

$d(x)=\sqrt{(x-0)^{2}+(\frac{x^{2}}{2}-5)^{2}}=\sqrt{x^{2}+\frac{x^{4}}{4}+25-5 x^{2}}=\sqrt{\frac{x^{4}}{4}-4 x^{2}+25}$

$\therefore d^{\prime}(x)=\frac{(x^{3}-8 x)}{2 \sqrt{\frac{x^{4}}{4}-4 x^{2}+25}}=\frac{(x^{3}-8 x)}{\sqrt{x^{4}-16 x^{2}+100}}$

Now, $d^{\prime}(x)=0 \Rightarrow x^{3}-8 x=0$

$\Rightarrow x(x^{2}-8)=0$

$\Rightarrow x=0, \pm 2 \sqrt{2}$

$ \begin{aligned} & \begin{aligned} & \text{ And, } d^{\prime \prime}(x)= \frac{\sqrt{x^{4}-16 x^{2}+100}(3 x^{2}-8)-(x^{3}-8 x) \cdot \frac{4 x^{3}-32 x}{2 \sqrt{x^{4}-16 x^{2}+100}}}{(x^{4}-16 x^{2}+100)} \\ &=\frac{(x^{4}-16 x^{2}+100)(3 x^{2}-8)-2(x^{3}-8 x)(x^{3}-8 x)}{(x^{4}-16 x^{2}+100)^{\frac{3}{2}}} \\ &=\frac{(x^{4}-16 x^{2}+100)(3 x^{2}-8)-2(x^{3}-8 x)^{2}}{(x^{4}-16 x^{2}+100)^{\frac{3}{2}}} \\ & \text{ When, } x=0 \text{, then } d^{\prime \prime}(x)=\frac{36(-8)}{6^{3}}<0 . \end{aligned} \\ & \text{ When, } x= \pm 2 \sqrt{2}, d^{\prime \prime}(x)>0 . \end{aligned} $

$\square$ By second derivative test, $d(x)$ is the minimum at $x= \pm 2 \sqrt{2}$.

When $x= \pm 2 \sqrt{2}, y=\frac{(2 \sqrt{2})^{2}}{2}=4$.

Hence, the point on the curve $x^{2}=2 y$ which is nearest to the point $(0,5)$ is $( \pm 2 \sqrt{2}, 4)$. The correct answer is $A$.

Solution

Let $f(x)=\frac{1-x+x^{2}}{1+x+x^{2}}$.

$ \begin{aligned} \therefore f^{\prime}(x) & =\frac{(1+x+x^{2})(-1+2 x)-(1-x+x^{2})(1+2 x)}{(1+x+x^{2})^{2}} \\ & =\frac{-1+2 x-x+2 x^{2}-x^{2}+2 x^{3}-1-2 x+x+2 x^{2}-x^{2}-2 x^{3}}{(1+x+x^{2})^{2}} \\ & =\frac{2 x^{2}-2}{(1+x+x^{2})^{2}}=\frac{2(x^{2}-1)}{(1+x+x^{2})^{2}} \end{aligned} $

$\therefore f^{\prime}(x)=0 \Rightarrow x^{2}=1 \Rightarrow x= \pm 1$

Now, $f^{\prime \prime}(x)=\frac{2[(1+x+x^{2})^{2}(2 x)-(x^{2}-1)(2)(1+x+x^{2})(1+2 x)]}{(1+x+x^{2})^{4}}$

$ =\frac{4(1+x+x^{2})[(1+x+x^{2}) x-(x^{2}-1)(1+2 x)]}{(1+x+x^{2})^{4}} $

$ \begin{aligned} & =\frac{4[x+x^{2}+x^{3}-x^{2}-2 x^{3}+1+2 x]}{(1+x+x^{2})^{3}} \\ & =\frac{4(1+3 x-x^{3})}{(1+x+x^{2})^{3}} \end{aligned} $

And, $f^{\prime \prime}(1)=\frac{4(1+3-1)}{(1+1+1)^{3}}=\frac{4(3)}{(3)^{3}}=\frac{4}{9}>0$

Also, $f^{\prime \prime}(-1)=\frac{4(1-3+1)}{(1-1+1)^{3}}=4(-1)=-4<0$

$\square$ By second derivative test, $f$ is the minimum at $x=1$ and the minimum value is given

by $f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$.

The correct answer is $D$.

Solution

Let $f(x)=[x(x-1)+1]^{\frac{1}{3}}$.

$\therefore f^{\prime}(x)=\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$

Then, we evaluate the value of $f$ at critical point $x=\frac{1}{2}$ and at the end points of the interval $[0,1]$ {i.e., at $x=0$ and $x=1}$.

$f(0)=[0(0-1)+1]^{\frac{1}{3}}=1$

$f(1)=[1(1-1)+1]^{\frac{1}{3}}=1$

$f(\frac{1}{2})=[\frac{1}{2}(\frac{-1}{2})+1]^{\frac{1}{3}}=(\frac{3}{4})^{\frac{1}{3}}$

Hence, we can conclude that the maximum value of $f$ in the interval $[0,1]$ is 1 .

The correct answer is $C$.

Solution

The given function is $f(x)=\frac{\log x}{x}$.

$ f^{\prime}(x)=\frac{x(\frac{1}{x})-\log x}{x^{2}}=\frac{1-\log x}{x^{2}} $

Now, $f^{\prime}(x)=0$

$\Rightarrow 1-\log x=0$ $\Rightarrow \log x=1$

$\Rightarrow \log x=\log e$

$\Rightarrow x=e$

Now, $f^{\prime \prime}(x)=\frac{x^{2}(-\frac{1}{x})-(1-\log x)(2 x)}{x^{4}}$

$ \begin{aligned} & =\frac{-x-2 x(1-\log x)}{x^{4}} \\ & =\frac{-3+2 \log x}{x^{3}} \end{aligned} $

Now, $f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^{3}}=\frac{-3+2}{e^{3}}=\frac{-1}{e^{3}}<0$

Therefore, by second derivative test, $f$ is the maximum at $x=e$.

Solution

Let $\triangle A B C$ be isosceles where $B C$ is the base of fixed length $b$.

Let the length of the two equal sides of $\triangle A B C$ be $a$.

Draw AD $\square BC$.

Now, in $\triangle A D C$, by applying the Pythagoras theorem, we have:

$AD=\sqrt{a^{2}-\frac{b^{2}}{4}}$

Area of triangle

$ (A)=\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}} $

The rate of change of the area with respect to time $(t)$ is given by,

$\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^{2}-\frac{b^{2}}{4}}} \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^{2}-b^{2}}} \frac{d a}{d t}$

It is given that the two equal sides of the triangle are decreasing at the rate of $3 cm$ per second.

$ \begin{aligned} & \\ & \square \frac{d a}{d t}=-3 cm / s \\ & \therefore \frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^{2}-b^{2}}} \end{aligned} $

Then, when $a=b$, we have:

$\frac{d A}{d t}=\frac{-3 b^{2}}{\sqrt{4 b^{2}-b^{2}}}=\frac{-3 b^{2}}{\sqrt{3 b^{2}}}=-\sqrt{3} b$

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $\sqrt{3} bm cm^{2} / s$.

Solution

$f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$

$\therefore f^{\prime}(x)=\frac{(2+\cos x)(4 \cos x-2-\cos x+x \sin x)-(4 \sin x-2 x-x \cos x)(-\sin x)}{(2+\cos x)^{2}}$

$=\frac{(2+\cos x)(3 \cos x-2+x \sin x)+\sin x(4 \sin x-2 x-x \cos x)}{(2+\cos x)^{2}}$

$=\frac{6 \cos x-4+2 x \sin x+3 \cos ^{2} x-2 \cos x+x \sin x \cos x+4 \sin ^{2} x-2 x \sin x-x \sin x \cos x}{(2+\cos x)^{2}}$

$=\frac{4 \cos x-4+3 \cos ^{2} x+4 \sin ^{2} x}{(2+\cos x)^{2}}$

$=\frac{4 \cos x-4+3 \cos ^{2} x+4-4 \cos ^{2} x}{(2+\cos x)^{2}}$

$=\frac{4 \cos x-\cos ^{2} x}{(2+\cos x)^{2}}=\frac{\cos x(4-\cos x)}{(2+\cos x)^{2}}$

Now, $f^{\prime}(x)=0$

$\Rightarrow \cos x=0$ or $\cos x=4$

But, $\cos x \neq 4$

$\square \cos x=0$

$\Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2}$

Now, $x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ divides $(0,2 \pi)$ into three disjoint intervals i.e.,

$(0, \frac{\pi}{2}),(\frac{\pi}{2}, \frac{3 \pi}{2})$, and $(\frac{3 \pi}{2}, 2 \pi)$.

In intervals $(0, \frac{\pi}{2})$ and $(\frac{3 \pi}{2}, 2 \pi), f^{\prime}(x)>0$.

Thus, $f(x)$ is increasing for $0<x<\frac{x}{2}$ and $\frac{3 \pi}{2}<x<2 \pi$.

In the interval $(\frac{\pi}{2}, \frac{3 \pi}{2}), f^{\prime}(x)<0$.

Thus, $f(x)$ is decreasing for $\frac{\pi}{2}<x<\frac{3 \pi}{2}$.

Solution

$f(x)=x^{3}+\frac{1}{x^{3}}$

$\therefore f^{\prime}(x)=3 x^{2}-\frac{3}{x^{4}}=\frac{3 x^{6}-3}{x^{4}}$

Then, $f^{\prime}(x)=0 \Rightarrow 3 x^{6}-3=0 \Rightarrow x^{6}=1 \Rightarrow x= \pm 1$

Now, the points $x=1$ and $x=-1$ divide the real line into three disjoint intervals

i.e., $(-\infty,-1),(-1,1)$, and $(1, \infty)$.

In intervals $(-\infty,-1)$ and $(1, \infty)$ i.e., when $x<-1$ and $x>1, f^{\prime}(x)>0$.

Thus, when $x<-1$ and $x>1, f$ is increasing.

In interval $(-1,1)$ i.e., when $-1<x<1, f^{\prime}(x)<0$.

Thus, when $-1<x<1, f$ is decreasing.

Solution

The given ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

Let the major axis be along the $x$-axis.

Let $A B C$ be the triangle inscribed in the ellipse where vertex $C$ is at $(a, 0)$.

Since the ellipse is symmetrical with respect to the $x$-axis and $y$-axis, we can assume the coordinates of $A$ to be $(-x_1, y_1)$ and the coordinates of $B$ to be $(-x_1,-y_1)$.

Now, we have $y_1= \pm \frac{b}{a} \sqrt{a^{2}-x_1^{2}}$.

$\square$ Coordinates of $A$ are $(-x_1, \frac{b}{a} \sqrt{a^{2}-x_1^{2}})$ and the coordinates of $B$ are $(x_1,-\frac{b}{a} \sqrt{a^{2}-x_1^{2}})$.

As the point $(x_1, y_1)$ lies on the ellipse, the area of triangle $ABC(A)$ is given by,

$ \begin{aligned} & A=\frac{1}{2}|a(\frac{2 b}{a} \sqrt{a^{2}-x_1^{2}})+(-x_1)(-\frac{b}{a} \sqrt{a^{2}-x_1^{2}})+(-x_1)(-\frac{b}{a} \sqrt{a^{2}-x_1^{2}})| \\ & \Rightarrow A=b \sqrt{a^{2}-x_1^{2}}+x_1 \frac{b}{a} \sqrt{a^{2}-x_1^{2}} \\ & \therefore \frac{d A}{d x_1}=\frac{-2 x_1 b}{2 \sqrt{a^{2}-x_1^{2}}}+\frac{b}{a} \sqrt{a^{2}-x_1^{2}}-\frac{2 b x_1^{2}}{a 2 \sqrt{a^{2}-x_1^{2}}} \\ & =\frac{b}{a \sqrt{a^{2}-x_1^{2}}}[-x_1 a+(a^{2}-x_1^{2})-x_1^{2}] \\ & =\frac{b(-2 x_1^{2}-x_1 a+a^{2})}{a \sqrt{a^{2}-x_1^{2}}} \end{aligned} $

Now, $\frac{d A}{d x_1}=0$

$\Rightarrow-2 x_1^{2}-x_1 a+a^{2}=0$

$\Rightarrow x_1=\frac{a \pm \sqrt{a^{2}-4(-2)(a^{2})}}{2(-2)}$

$=\frac{a \pm \sqrt{9 a^{2}}}{-4}$

$=\frac{a \pm 3 a}{-4}$

$\Rightarrow x_1=-a, \frac{a}{2}$

But, $x_1$ cannot be equal to $a$. $\therefore x_1=\frac{a}{2} \Rightarrow y_1=\frac{b}{a} \sqrt{a^{2}-\frac{a^{2}}{4}}=\frac{b a}{2 a} \sqrt{3}=\frac{\sqrt{3} b}{2}$

Now, $\frac{d^{2} A}{d x_1^{2}}=\frac{b}{a}{\frac{\sqrt{a^{2}-x_1^{2}}(-4 x_1-a)-(-2 x_1^{2}-x_1 a+a^{2}) \frac{(-2 x_1)}{2 \sqrt{a^{2}-x_1^{2}}}}{a^{2}-x_1^{2}}}$

$ =\frac{b}{a}{\frac{(a^{2}-x_1^{2})(-4 x_1-a)+x_1(-2 x_1^{2}-x_1 a+a^{2})}{(a^{2}-x_1^{2})^{\frac{3}{2}}}} $

$ =\frac{b}{a}{\frac{2 x^{3}-3 a^{2} x-a^{3}}{(a^{2}-x_1^{2})^{\frac{3}{2}}}} $

Also, when $x_1=\frac{a}{2}$, then

$ \begin{aligned} \frac{d^{2} A}{d x_1^{2}} & =\frac{b}{a}{\frac{2 \frac{a^{3}}{8}-3 \frac{a^{3}}{2}-a^{3}}{(\frac{3 a^{2}}{4})^{\frac{3}{2}}}}=\frac{b}{a}{\frac{\frac{a^{3}}{4}-\frac{3}{2} a^{3}-a^{3}}{(\frac{3 a^{2}}{4})^{\frac{3}{2}}}} \\ & =-\frac{b}{a}{\frac{\frac{9}{4} a^{3}}{(\frac{3 a^{2}}{4})^{\frac{3}{2}}}}<0 \end{aligned} $

Thus, the area is the maximum when $x_1=\frac{a}{2}$.

$\square$ Maximum area of the triangle is given by,

$ \begin{aligned} A & =b \sqrt{a^{2}-\frac{a^{2}}{4}}+(\frac{a}{2}) \frac{b}{a} \sqrt{a^{2}-\frac{a^{2}}{4}} \\ & =a b \frac{\sqrt{3}}{2}+(\frac{a}{2}) \frac{b}{a} \times \frac{a \sqrt{3}}{2} \\ & =\frac{a b \sqrt{3}}{2}+\frac{a b \sqrt{3}}{4}=\frac{3 \sqrt{3}}{4} a b \end{aligned} $

Solution

Let $l, b$, and $h$ represent the length, breadth, and height of the tank respectively.

Then, we have height $(h)=2 m$

Volume of the tank $=8 m^{3}$

Volume of the tank $=l \times b \times h$

$\square 8=l \times b \times 2$

$\Rightarrow l b=4 \Rightarrow b=\frac{4}{l}$

Now, area of the base $=I b=4$

Area of the 4 walls $(A)=2 h(I+b)$

$\therefore A=4(l+\frac{4}{l})$

$\Rightarrow \frac{d A}{d l}=4(1-\frac{4}{l^{2}})$

Now, $\frac{d A}{d l}=0$

$\Rightarrow 1-\frac{4}{l^{2}}=0$

$\Rightarrow l^{2}=4$

$\Rightarrow l= \pm 2$

However, the length cannot be negative.

Therefore, we have $I=4$.

$\therefore b=\frac{4}{l}=\frac{4}{2}=2$

Now, $\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}}$

When $l=2, \frac{d^{2} A}{d l^{2}}=\frac{32}{8}=4>0$.

Thus, by second derivative test, the area is the minimum when $I=2$.

We have $I=b=h=2$.

$\square$ Cost of building the base $=Rs 70 \times(I b)=Rs 70(4)=Rs 280$

Cost of building the walls $=Rs 2 h(I+b) \times 45=Rs 90(2)(2+2)$

$=Rs 8(90)=Rs 720$

Required total cost $=Rs(280+720)=Rs 1000$

Hence, the total cost of the tank will be Rs 1000.

Solution

Let $r$ be the radius of the circle and $a$ be the side of the square.

Then, we have:

$2 \pi r+4 a=k$ (where $k$ is constant)

$\Rightarrow a=\frac{k-2 \pi r}{4}$

The sum of the areas of the circle and the square $(A)$ is given by,

$A=\pi r^{2}+a^{2}=\pi r^{2}+\frac{(k-2 \pi r)^{2}}{16}$

$\therefore \frac{d A}{d r}=2 \pi r+\frac{2(k-2 \pi r)(-2 \pi)}{16}=2 \pi r-\frac{\pi(k-2 \pi r)}{4}$

Now, $\frac{d A}{d r}=0$

$\Rightarrow 2 \pi r=\frac{\pi(k-2 \pi r)}{4}$

$8 r=k-2 \pi r$

$\Rightarrow(8+2 \pi) r=k$

$\Rightarrow r=\frac{k}{8+2 \pi}=\frac{k}{2(4+\pi)}$

Now, $\frac{d^{2} A}{d r^{2}}=2 \pi+\frac{\pi^{2}}{2}>0$

$\therefore$ When $r=\frac{k}{2(4 \pi)}, \frac{d^{2} A}{d r^{2}}>0$.

When $r=\frac{k}{2(4 \pi)}, a=\frac{k-2 \pi[\frac{k}{2(4 \pi)}]}{4}=\frac{k(4 \pi) \pi k}{4(\pi)}=\frac{4 k}{4(\pi)}=\frac{k}{\pi}=2 r$.

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Solution

Let $x$ and $y$ be the length and breadth of the rectangular window.

Radius of the semicircular opening $=\frac{x}{2}$

It is given that the perimeter of the window is $10 m$.

$\therefore x+2 y+\frac{\pi x}{2}=10$

$\Rightarrow x(1+\frac{\pi}{2})+2 y=10$

$\Rightarrow 2 y=10-x(1+\frac{\pi}{2})$

$\Rightarrow y=5-x(\frac{1 \pi}{2}+\frac{-}{4})$

$\square$ Area of the window $(A)$ is given by,

$ \begin{aligned} A & =x y+\frac{\pi}{2}(\frac{x}{2})^{2} \\ & =x[5-x(\frac{1 \pi}{2}+\frac{\pi}{4})]+\frac{x^{2}}{8} \\ & =5 x-x^{2}(\frac{1 \pi}{2}+\frac{\pi}{4})+\frac{\pi}{8} x^{2} \\ \therefore & \frac{d A}{d x}=5-2 x(\frac{1 \pi}{2}+\frac{\pi}{4})+\frac{\pi}{4} x \\ & =5-x(1+\frac{\pi}{2})+\frac{\pi}{4} x \\ \therefore & \frac{d^{2} A}{d x^{2}}=-(1+\frac{\pi}{2})+\frac{\pi}{4}=-1-\frac{\pi}{4} \end{aligned} $

Now, $\frac{d A}{d x}=0$

$\Rightarrow 5-x(1+\frac{\pi}{2})+\frac{\pi}{4} x=0$

$\Rightarrow 5-x-\frac{\pi}{4} x=0$

$\Rightarrow x(1+\frac{\pi}{4})=5$

$\Rightarrow x=\frac{5}{(1+\frac{\pi}{4})}=\frac{20}{\pi+4}$

Thus, when $x=\frac{20}{\pi+4}$ then $\frac{d^{2} A}{d x^{2}}<0$.

Therefore, by second derivative test, the area is the maximum when length $x=\frac{20}{\pi+4} m$. Now,

$y=5-\frac{20}{\pi+4}(\frac{2+\pi}{4})=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4} m$

Hence, the required dimensions of the window to admit maximum light is given

by length $=\frac{20}{\pi+4} m$ and breadth $=\frac{10}{\pi+4} m$.

Solution

Let $\triangle A B C$ be right-angled at $B$. Let $A B=x$ and $B C=y$.

Let $P$ be a point on the hypotenuse of the triangle such that $P$ is at a distance of $a$ and $b$ from the sides $A B$ and $B C$ respectively.

Let $\square C=\theta$.

We have,

$AC=\sqrt{x^{2}+y^{2}}$

Now,

$PC=b cosec \theta$

And, $AP=a \sec \theta$

$\square AC=AP+PC$

$\square AC=b cosec \theta+a \sec \theta \ldots$ (1)

$\therefore \frac{d(AC)}{d \theta}=-b cosec \theta \cot \theta+a \sec \theta \tan \theta$

$\therefore \frac{d(AC)}{d \theta}=0$

$\Rightarrow a \sec \theta \tan \theta=b cosec \theta \cot \theta$

$\Rightarrow \frac{a}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta}=\frac{b}{\sin \theta} \frac{\cos \theta}{\sin \theta}$

$\Rightarrow a \sin ^{3} \theta=b \cos ^{3} \theta$

$\Rightarrow(a)^{\frac{1}{3}} \sin \theta=(b)^{\frac{1}{3}} \cos \theta$

$\Rightarrow \tan \theta=(\frac{b}{a})^{\frac{1}{3}}$

$\therefore \sin \theta=\frac{(b)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}$ and $\cos \theta=\frac{(a)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}$

It can be clearly shown that $\frac{d^{2}(AC)}{d \theta^{2}}<0$ when $\tan \theta=(\frac{b}{a})^{\frac{1}{3}}$.

Therefore, by second derivative test, the length of the hypotenuse is the maximum when

$\tan \theta=(\frac{b}{a})^{\frac{1}{3}}$.

Now, when $\tan \theta=(\frac{b}{a})^{\frac{1}{3}}$, we have:

$ \begin{aligned} AC & =\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}+\frac{a \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}} \\ & =\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}(b^{\frac{2}{3}}+a^{\frac{2}{3}}) \\ & =(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}} \end{aligned} $

Hence, the maximum length of the hypotenuses is $(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$.

Solution

The given function is $f(x)=(x-2)^{4}(x+1)^{3}$.

$ \begin{aligned} \therefore f^{\prime}(x) & =4(x-2)^{3}(x+1)^{3}+3(x+1)^{2}(x-2)^{4} \\ & =(x-2)^{3}(x+1)^{2}[4(x+1)+3(x-2)] \\ & =(x-2)^{3}(x+1)^{2}(7 x-2) \end{aligned} $

Now, $f^{\prime}(x)=0 \Rightarrow x=-1$ and $x=\frac{2}{7}$ or $x=2$

Now, for values of $x$ close to $\frac{2}{7}$ and to the left of $\frac{2}{7}, f^{\prime}(x)>0$. Also, for values of $x$ close to $\frac{2}{7}$ and to the right of $\frac{2}{7}, f^{\prime}(x)<0$.

Thus, $x=\frac{2}{7}$ is the point of local maxima.

Now, for values of $x$ close to 2 and to the left of $2, f^{\prime}(x)<0$. Also, for values of $x$ close to 2 and to the right of $2, f^{\prime}(x)>0$.

Thus, $x=2$ is the point of local minima.

Now, as the value of $x$ varies through $-1, f^{\prime}(x)$ does not changes its sign.

Thus, $x=-1$ is the point of inflexion.

Solution

$ \begin{aligned} f(x) & =\cos ^{2} x+\sin x \\ f^{\prime}(x) & =2 \cos x(-\sin x)+\cos x \\ & =-2 \sin x \cos x+\cos x \end{aligned} $

Now, $f^{\prime}(x)=0$

$\Rightarrow 2 \sin x \cos x=\cos x \Rightarrow \cos x(2 \sin x-1)=0$

$\Rightarrow \sin x=\frac{1}{2}$ or $\cos x=0$

$\Rightarrow x=\frac{\pi}{6}$, or $\frac{\pi}{2}$ as $x \in[0, \pi]$

Now, evaluating the value of $f$ at critical points $x=\frac{\pi}{2}$ and $x=\frac{\pi}{6}$ and at the end points of

the interval $[0, \pi]$ (i.e., at $x=0$ and $x=\pi$ ), we have: $f(\frac{\pi}{6})=\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6}=(\frac{\sqrt{3}}{2})^{2}+\frac{1}{2}=\frac{5}{4}$

$f(0)=\cos ^{2} 0+\sin 0=1+0=1$

$f(\pi)=\cos ^{2} \pi+\sin \pi=(-1)^{2}+0=1$

$f(\frac{\pi}{2})=\cos ^{2} \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1$

Hence, the absolute maximum value of $f$ is $\frac{5}{4}$ occurring at $x=\frac{\pi}{6}$ and the absolute minimum value of $f$ is 1 occurring at $x=0, \frac{\pi}{2}$, and $\pi$.

Solution

A sphere of fixed radius $(r)$ is given.

Let $R$ and $h$ be the radius and the height of the cone respectively.

The volume $(V)$ of the cone is given by,

$V=\frac{1}{3} \pi R^{2} h$

Now, from the right triangle $B C D$, we have:

$BC=\sqrt{r^{2}-R^{2}}$

$\square h=r+\sqrt{r^{2}-R^{2}}$

$ \begin{aligned} & \therefore V=\frac{1}{3} \pi R^{2}(r+\sqrt{r^{2}-R^{2}})=\frac{1}{3} \pi R^{2} r+\frac{1}{3} \pi R^{2} \sqrt{r^{2}-R^{2}} \\ & \begin{aligned} \therefore \frac{d V}{d R} & =\frac{2}{3} \pi R r+\frac{2 \pi}{3} \pi R \sqrt{r^{2}-R^{2}}+\frac{R^{2}}{3} \cdot \frac{(-2 R)}{2 \sqrt{r^{2}-R^{2}}} \\ & =\frac{2}{3} \pi R r+\frac{2 \pi}{3} \pi R \sqrt{r^{2}-R^{2}}-\frac{R^{3}}{3 \sqrt{r^{2}-R^{2}}} \\ & =\frac{2}{3} \pi R r+\frac{2 \pi R(r^{2}-R^{2})-\pi R^{3}}{3 \sqrt{r^{2}-R^{2}}} \\ & =\frac{2}{3} \pi R r+\frac{2 \pi R r^{2}-3 \pi R^{3}}{3 \sqrt{r^{2}-R^{2}}} \end{aligned} \end{aligned} $

Now, $\frac{d V}{d R^{2}}=0$

$\Rightarrow \frac{2 \pi r R}{3}=\frac{3 \pi R^{3}-2 \pi R r^{2}}{3 \sqrt{r^{2}-R^{2}}}$

$\Rightarrow 2 r \sqrt{r^{2}-R^{2}}=3 R^{2}-2 r^{2}$

$\Rightarrow 4 r^{2}(r^{2}-R^{2})=(3 R^{2}-2 r^{2})^{2}$

$\Rightarrow 4 r^{4}-4 r^{2} R^{2}=9 R^{4}+4 r^{4}-12 R^{2} r^{2}$

$\Rightarrow 9 R^{4}-8 r^{2} R^{2}=0$

$\Rightarrow 9 R^{2}=8 r^{2}$

$\Rightarrow R^{2}=\frac{8 r^{2}}{9}$

Now, $\frac{d^{2} V}{d R^{2}}=\frac{2 \pi r}{3}+\frac{3 \sqrt{r^{2}-R^{2}}(2 \pi r^{2}-9 \pi R^{2})-(2 \pi R r^{2}-3 \pi R^{3})(-6 R) \frac{1}{2 \sqrt{r^{2}-R^{2}}}}{9(r^{2}-R^{2})}$

$ =\frac{2 \pi r}{3}+\frac{3 \sqrt{r^{2}-R^{2}}(2 \pi r^{2}-9 \pi R^{2})+(2 \pi R r^{2}-3 \pi R^{3})(3 R) \frac{1}{2 \sqrt{r^{2}-R^{2}}}}{9(r^{2}-R^{2})} $

Now, when $R^{2}=\frac{8 r^{2}}{9}$, it can be shown that $\frac{d^{2} V}{d R^{2}}<0$.

$\square$ The volume is the maximum when

$ R^{2}=\frac{8 r^{2}}{9} $

When $R^{2}=\frac{8 r^{2}}{9}$, height of the cone $=r+\sqrt{r^{2}-\frac{8 r^{2}}{9}}=r+\sqrt{\frac{r^{2}}{9}}=r+\frac{r}{3}=\frac{4 r}{3}$.

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$.

Solution

Let us take any 2 points $\mathrm{c}_1$ and $\mathrm{c}_2$ such that

${\mathrm{c} _1, \mathrm{c} _2} \in(\mathrm{a}, \mathrm{b})$ and

$\mathrm{c} _2=\mathrm{c} _1+\mathrm{h}$,

where $\mathrm{h} \rightarrow \mathrm{O}$

Now, $f^{\prime}\left(c _1\right)=\lim _{h \rightarrow 0} \frac{f\left(c _1+h\right)-f\left(c _1\right)}{h}=\frac{f\left(c _2\right)-f\left(c _1\right)}{c _2-c _1}$

Now, it is given that $\mathrm{f}^{\prime}(\mathrm{x})>0 \quad \forall \mathrm{x} \in(\mathrm{a}, \mathrm{b})$. Therefore, $\mathrm{f}^{\prime}\left(\mathrm{c}_1\right)>0$

$\Rightarrow \frac{\mathrm{f}\left(\mathrm{c}_2\right)-\mathrm{f}\left(\mathrm{c}_1\right)}{\mathrm{c}_2-\mathrm{c}_1}>\mathrm{O}$

From the above fraction, we can conclude that:

1. For $\mathrm{c}_2>\mathrm{c}_1, \mathrm{f}\left(\mathrm{c}_2\right)>\mathrm{f}\left(\mathrm{c}_1\right)$

2. For $c_1>c_2, f\left(c_1\right)>f\left(c_2\right)$

Hence, $\mathrm{f}(\mathrm{x})$ is an increasing function in $(\mathrm{a}, \mathrm{b})$.

Solution

A sphere of fixed radius $(R)$ is given.

Let $r$ and $h$ be the radius and the height of the cylinder respectively.

From the given figure, we have $h=2 \sqrt{R^{2}-r^{2}}$.

The volume $(V)$ of the cylinder is given by,

$ \begin{aligned} & V=\pi r^{2} h=2 \pi r^{2} \sqrt{R^{2}-r^{2}} \\ & \begin{aligned} \therefore \frac{d V}{d r} & =4 \pi r \sqrt{R^{2}-r^{2}}+\frac{2 \pi r^{2}(-2 r)}{2 \sqrt{R^{2}-r^{2}}} \\ & =4 \pi r \sqrt{R^{2}-r^{2}}-\frac{2 \pi r^{3}}{\sqrt{R^{2}-r^{2}}} \\ & =\frac{4 \pi r(R^{2}-r^{2})-2 \pi r^{3}}{\sqrt{R^{2}-r^{2}}} \\ & =\frac{4 \pi r R^{2}-6 \pi r^{3}}{\sqrt{R^{2}-r^{2}}} \end{aligned} \end{aligned} $

Now, $\frac{d V}{d r}=0 \Rightarrow 4 \pi r R^{2}-6 \pi r^{3}=0$

$\Rightarrow r^{2}=\frac{2 R^{2}}{3}$

Now, $\frac{d^{2} V}{d r^{2}}=\frac{\sqrt{R^{2}-r^{2}}(4 \pi R^{2}-18 \pi r^{2})-(4 \pi r R^{2}-6 \pi r^{3}) \frac{(-2 r)}{2 \sqrt{R^{2}-r^{2}}}}{(R^{2}-r^{2})}$

$ \begin{aligned} & =\frac{(R^{2}-r^{2})(4 \pi R^{2}-18 \pi r^{2})+r(4 \pi r R^{2}-6 \pi r^{3})}{(R^{2}-r^{2})^{\frac{3}{2}}} \\ & =\frac{4 \pi R^{4}-22 \pi r^{2} R^{2}+12 \pi r^{4}+4 \pi r^{2} R^{2}}{(R^{2}-r^{2})^{\frac{3}{2}}} \end{aligned} $

Now, it can be observed that at $r^{2}=\frac{2 R^{2}}{3}, \frac{d^{2} V}{d r^{2}}<0$.

$\square$ The volume is the maximum when $r^{2}=\frac{2 R^{2}}{3}$.

When $r^{2}=\frac{2 R^{2}}{3}$, the height of the cylinder is $2 \sqrt{R^{2}-\frac{2 R^{2}}{3}}=2 \sqrt{\frac{R^{2}}{3}}=\frac{2 R}{\sqrt{3}}$.

Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2 R}{\sqrt{3}}$.

Solution

The given right circular cone of fixed height $(h)$ and semi-vertical angle ( $a$ ) can be drawn as:

Here, a cylinder of radius $R$ and height $H$ is inscribed in the cone.

Then, $\square GAO=a, OG=r, OA=h, OE=R$, and $CE=H$.

We have,

$r=h \tan a$

Now, since $\triangle AOG$ is similar to $\triangle CEG$, we have:

$\frac{AO}{OG}=\frac{CE}{EG}$

$\Rightarrow \frac{h}{r}=\frac{H}{r-R} \quad[EG=OG-OE]$

$\Rightarrow H=\frac{h}{r}(r-R)=\frac{h}{h \tan \alpha}(h \tan \alpha-R)=\frac{1}{\tan \alpha}(h \tan \alpha-R)$

Now, the volume $(V)$ of the cylinder is given by,

$V=\pi R^{2} H=\frac{\pi R^{2}}{\tan \alpha}(h \tan \alpha-R)=\pi R^{2} h-\frac{\pi R^{3}}{\tan \alpha}$

$\therefore \frac{d V}{d R}=2 \pi R h-\frac{3 \pi R^{2}}{\tan \alpha}$

Now, $\frac{d V}{d R}=0$

$\Rightarrow 2 \pi R h=\frac{3 \pi R^{2}}{\tan \alpha}$

$\Rightarrow 2 h \tan \alpha=3 R$

$\Rightarrow R=\frac{2 h}{3} \tan \alpha$

Now, $\frac{d^{2} V}{d R^{2}}=2 \pi h-\frac{6 \pi R}{\tan \alpha}$

And, for $R=\frac{2 h}{3} \tan \alpha$, we have:

$\frac{d^{2} V}{d R^{2}}=2 \pi h-\frac{6 \pi}{\tan \alpha}(\frac{2 h}{3} \tan \alpha)=2 \pi h-4 \pi h=-2 \pi h<0$

$\square$ By second derivative test, the volume of the cylinder is the greatest when

$R=\frac{2 h}{3} \tan \alpha$.

When $R=\frac{2 h}{3} \tan \alpha, H=\frac{1}{\tan \alpha}(h \tan \alpha-\frac{2 h}{3} \tan \alpha)=\frac{1}{\tan \alpha}(\frac{h \tan \alpha}{3})=\frac{h}{3}$.

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Now, the maximum volume of the cylinder can be obtained as:

$\pi(\frac{2 h}{3} \tan \alpha)^{2}(\frac{h}{3})=\pi(\frac{4 h^{2}}{9} \tan ^{2} \alpha)(\frac{h}{3})=\frac{4}{27} \pi h^{3} \tan ^{2} \alpha$

Hence, the given result is proved.

Solution

Let $r$ be the radius of the cylinder.

Then, volume $(V)$ of the cylinder is given by,

$V=\pi(\text{ radius })^{2} \times$ height

$ \begin{matrix} =\pi(10)^{2} h & (\text{ radius }=10 m) \\ =100 \pi h \end{matrix} $

Differentiating with respect to time $t$, we have:

$\frac{d V}{d t}=100 \pi \frac{d h}{d t}$

The tank is being filled with wheat at the rate of 314 cubic metres per hour.

$ \square \frac{d V}{d t}=314 m^{3} / h $

Thus, we have:

$ \begin{aligned} & 314=100 \pi \frac{d h}{d t} \\ & \Rightarrow \frac{d h}{d t}=\frac{314}{100(3.14)}=\frac{314}{314}=1 \end{aligned} $

Hence, the depth of wheat is increasing at the rate of $1 m / h$.

The correct answer is $A$.